\displaystyle \textbf{Exercise - 12(A)}

\displaystyle \textbf{Question 1: }\text{In }\triangle ABC,\ M\text{ is the mid-point of }AB\text{ and a straight line}
\displaystyle \text{through }M\text{ and parallel to }BC\text{ cuts }AC\text{ at }N.\text{ Find the lengths of }AN
\displaystyle \text{and }MN,\text{ if }BC=7\text{ cm and }AC=5\text{ cm.}
\displaystyle \textbf{Answer:} \displaystyle \text{Since }M\text{ is the mid-point of }AB\text{ and }MN\parallel BC,
\displaystyle \text{by the converse of the Mid-point Theorem, }N\text{ is the mid-point of }AC.
\displaystyle \therefore AN=\frac{1}{2}AC=\frac{1}{2}\times5=2.5\text{ cm}.
\displaystyle \text{Also, }M\text{ and }N\text{ are the mid-points of }AB\text{ and }AC,\text{ respectively}.
\displaystyle \text{By the Mid-point Theorem, }MN=\frac{1}{2}BC.
\displaystyle \therefore MN=\frac{1}{2}\times7=3.5\text{ cm}.
\displaystyle {\therefore AN=2.5\text{ cm and }MN=3.5\text{ cm}.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Prove that the figure obtained by joining the mid-points of the adjacent}
\displaystyle \text{sides of a rectangle is a rhombus.}
\displaystyle \textbf{Answer:} \displaystyle \text{Let }P,\ Q,\ R\text{ and }S\text{ be the mid-points of }AB,\ BC,\ CD\text{ and }DA,
\displaystyle \text{respectively, of rectangle }ABCD.
\displaystyle \text{Join }PQ,\ QR,\ RS\text{ and }SP.
\displaystyle \text{In }\triangle ABC,\ P\text{ and }Q\text{ are the mid-points of }AB\text{ and }BC.
\displaystyle \therefore PQ=\frac{1}{2}AC\qquad\text{(By the Mid-point Theorem).}
\displaystyle \text{In }\triangle BCD,\ Q\text{ and }R\text{ are the mid-points of }BC\text{ and }CD.
\displaystyle \therefore QR=\frac{1}{2}BD\qquad\text{(By the Mid-point Theorem).}
\displaystyle \text{In }\triangle CDA,\ R\text{ and }S\text{ are the mid-points of }CD\text{ and }DA.
\displaystyle \therefore RS=\frac{1}{2}AC\qquad\text{(By the Mid-point Theorem).}
\displaystyle \text{In }\triangle DAB,\ S\text{ and }P\text{ are the mid-points of }DA\text{ and }AB.
\displaystyle \therefore SP=\frac{1}{2}BD\qquad\text{(By the Mid-point Theorem).}
\displaystyle \text{Since the diagonals of a rectangle are equal, }AC=BD.
\displaystyle \therefore \frac{1}{2}AC=\frac{1}{2}BD.
\displaystyle \therefore PQ=QR=RS=SP.
\displaystyle {\therefore \text{Quadrilateral }PQRS\text{ is a rhombus. Hence proved.}}
\displaystyle \\

\displaystyle \textbf{Question 3: }D,\ E\text{ and }F\text{ are the mid-points of the sides }AB,\ BC\text{ and }CA
\displaystyle \text{of an isosceles }\triangle ABC\text{ in which }AB=BC.\text{ Prove that }\triangle DEF\text{ is also}
\displaystyle \text{isosceles.}
\displaystyle \textbf{Answer:} \displaystyle \text{In }\triangle ABC,\ E\text{ and }F\text{ are the mid-points of }BC\text{ and }CA.
\displaystyle \therefore EF=\frac{1}{2}AB\qquad\text{(By the Mid-point Theorem).}
\displaystyle \text{Also, }D\text{ and }F\text{ are the mid-points of }AB\text{ and }AC.
\displaystyle \therefore DF=\frac{1}{2}BC\qquad\text{(By the Mid-point Theorem).}
\displaystyle \text{But }AB=BC\qquad\text{(Given).}
\displaystyle \therefore \frac{1}{2}AB=\frac{1}{2}BC.
\displaystyle \therefore EF=DF.
\displaystyle {\therefore \triangle DEF\text{ is an isosceles triangle. Hence proved.}}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{The following figure shows a trapezium }ABCD\text{ in which }AB\parallel DC.
\displaystyle P\text{ is the mid-point of }AD\text{ and }PR\parallel AB.\text{ Prove that:}
\displaystyle PR=\frac{1}{2}(AB+CD). \displaystyle \textbf{Answer:}
\displaystyle \text{Draw the diagonal }DB,\text{ which intersects }PR\text{ at }Q.
\displaystyle \text{In }\triangle ADB,\ P\text{ is the mid-point of }AD\text{ and }PQ\parallel AB.
\displaystyle \text{By the converse of the Mid-point Theorem, }Q\text{ is the mid-point of }DB.
\displaystyle \therefore PQ=\frac{1}{2}AB\qquad\text{(By the Mid-point Theorem).}
\displaystyle \text{In }\triangle DBC,\ Q\text{ is the mid-point of }DB\text{ and }QR\parallel DC.
\displaystyle \text{By the converse of the Mid-point Theorem, }R\text{ is the mid-point of }BC.
\displaystyle \therefore QR=\frac{1}{2}DC\qquad\text{(By the Mid-point Theorem).}
\displaystyle \text{Since }Q\text{ lies on }PR,\ PR=PQ+QR.
\displaystyle \therefore PR=\frac{1}{2}AB+\frac{1}{2}DC.
\displaystyle \therefore PR=\frac{1}{2}(AB+DC).
\displaystyle {\therefore PR=\frac{1}{2}(AB+CD).\text{ Hence proved.}}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{The figure given below shows a trapezium }ABCD.\ M\text{ and }N\text{ are the}
\displaystyle \text{mid-points of the non-parallel sides }AD\text{ and }BC,\text{ respectively. Find:}
\displaystyle \text{(i) }MN,\text{ if }AB=11\text{ cm and }DC=8\text{ cm.}
\displaystyle \text{(ii) }AB,\text{ if }DC=20\text{ cm and }MN=27\text{ cm.}
\displaystyle \text{(iii) }DC,\text{ if }MN=15\text{ cm and }AB=23\text{ cm.} \displaystyle \textbf{Answer:}
\displaystyle \text{Since }M\text{ and }N\text{ are the mid-points of the non-parallel sides of the trapezium,}
\displaystyle MN=\frac{1}{2}(AB+DC).
\displaystyle \text{(i) }MN=\frac{1}{2}(11+8)=\frac{19}{2}=9.5\text{ cm}.
\displaystyle {\therefore MN=9.5\text{ cm}.}
\displaystyle \text{(ii) }27=\frac{1}{2}(AB+20).
\displaystyle \therefore 54=AB+20.
\displaystyle \therefore AB=54-20=34\text{ cm}.
\displaystyle {\therefore AB=34\text{ cm}.}
\displaystyle \text{(iii) }15=\frac{1}{2}(23+DC).
\displaystyle \therefore 30=23+DC.
\displaystyle \therefore DC=30-23=7\text{ cm}.
\displaystyle {\therefore DC=7\text{ cm}.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{The diagonals of a quadrilateral intersect at right angles. Prove that the}
\displaystyle \text{figure obtained by joining the mid-points of the adjacent sides of the quadrilateral}
\displaystyle \text{is a rectangle.}
\displaystyle \textbf{Answer:} \displaystyle \text{Let }P,\ Q,\ R\text{ and }S\text{ be the mid-points of }AB,\ BC,\ CD\text{ and }DA,
\displaystyle \text{respectively, of quadrilateral }ABCD,\text{ where }AC\perp BD.
\displaystyle \text{Join }PQ,\ QR,\ RS\text{ and }SP.
\displaystyle \text{In }\triangle ABC,\ P\text{ and }Q\text{ are the mid-points of }AB\text{ and }BC.
\displaystyle \therefore PQ\parallel AC\qquad\text{(By the Mid-point Theorem).}
\displaystyle \text{In }\triangle BCD,\ Q\text{ and }R\text{ are the mid-points of }BC\text{ and }CD.
\displaystyle \therefore QR\parallel BD\qquad\text{(By the Mid-point Theorem).}
\displaystyle \text{In }\triangle CDA,\ R\text{ and }S\text{ are the mid-points of }CD\text{ and }DA.
\displaystyle \therefore RS\parallel AC\qquad\text{(By the Mid-point Theorem).}
\displaystyle \text{In }\triangle DAB,\ S\text{ and }P\text{ are the mid-points of }DA\text{ and }AB.
\displaystyle \therefore SP\parallel BD\qquad\text{(By the Mid-point Theorem).}
\displaystyle \therefore PQ\parallel RS\text{ and }QR\parallel SP.
\displaystyle \therefore PQRS\text{ is a parallelogram}.
\displaystyle \text{Also, }AC\perp BD,\ PQ\parallel AC\text{ and }QR\parallel BD.
\displaystyle \therefore PQ\perp QR,\text{ i.e., }\angle PQR=90^\circ.
\displaystyle \text{Thus, }PQRS\text{ is a parallelogram having one angle equal to }90^\circ.
\displaystyle {\therefore PQRS\text{ is a rectangle. Hence proved.}}
\displaystyle \\

\displaystyle \textbf{Question 7: }L\text{ and }M\text{ are the mid-points of sides }AB\text{ and }DC,\text{ respectively,}
\displaystyle \text{of parallelogram }ABCD.\text{ Prove that the segments }DL\text{ and }BM\text{ trisect}
\displaystyle \text{the diagonal }AC.
\displaystyle \textbf{Answer:} \displaystyle \text{Let }DL\text{ and }BM\text{ intersect the diagonal }AC\text{ at }P\text{ and }Q,
\displaystyle \text{respectively}.
\displaystyle \text{Since }L\text{ is the mid-point of }AB,\ AL=\frac{1}{2}AB.
\displaystyle \text{Also, }AB=DC\qquad\text{(Opposite sides of a parallelogram).}
\displaystyle \therefore AL=\frac{1}{2}DC.
\displaystyle \text{In }\triangle APL\text{ and }\triangle CPD,
\displaystyle \angle APL=\angle CPD\qquad\text{(Vertically opposite angles),}
\displaystyle \angle PAL=\angle PCD\qquad\text{(Alternate angles, since }AB\parallel DC\text{).}
\displaystyle \therefore \triangle APL\sim\triangle CPD\qquad\text{(A.A. similarity).}
\displaystyle \therefore \frac{AP}{PC}=\frac{AL}{CD}=\frac{1}{2}.
\displaystyle \therefore AP:PC=1:2.
\displaystyle \therefore AP=\frac{1}{3}AC.
\displaystyle \text{Since }M\text{ is the mid-point of }DC,\ CM=\frac{1}{2}DC.
\displaystyle \text{But }DC=AB.
\displaystyle \therefore CM=\frac{1}{2}AB.
\displaystyle \text{In }\triangle ABQ\text{ and }\triangle CMQ,
\displaystyle \angle AQB=\angle CQM\qquad\text{(Vertically opposite angles),}
\displaystyle \angle BAQ=\angle MCQ\qquad\text{(Alternate angles, since }AB\parallel DC\text{).}
\displaystyle \therefore \triangle ABQ\sim\triangle CMQ\qquad\text{(A.A. similarity).}
\displaystyle \therefore \frac{AQ}{QC}=\frac{AB}{CM}=2.
\displaystyle \therefore AQ:QC=2:1.
\displaystyle \therefore QC=\frac{1}{3}AC.
\displaystyle \text{Now, }PQ=AC-AP-QC.
\displaystyle \therefore PQ=AC-\frac{1}{3}AC-\frac{1}{3}AC=\frac{1}{3}AC.
\displaystyle \therefore AP=PQ=QC.
\displaystyle {\therefore DL\text{ and }BM\text{ trisect the diagonal }AC.\text{ Hence proved.}}
\displaystyle \\

\displaystyle \textbf{Question 8: }ABCD\text{ is a quadrilateral in which }AD=BC.\ E,\ F,\ G\text{ and }H
\displaystyle \text{are the mid-points of }AB,\ BD,\ CD\text{ and }AC,\text{ respectively. Prove that }EFGH
\displaystyle \text{is a rhombus.} \displaystyle \textbf{Answer:}
\displaystyle \text{In }\triangle ABD,\ E\text{ and }F\text{ are the mid-points of }AB\text{ and }BD.
\displaystyle \therefore EF=\frac{1}{2}AD\qquad\text{(By the Mid-point Theorem).}
\displaystyle \text{In }\triangle BCD,\ F\text{ and }G\text{ are the mid-points of }BD\text{ and }CD.
\displaystyle \therefore FG=\frac{1}{2}BC\qquad\text{(By the Mid-point Theorem).}
\displaystyle \text{In }\triangle ACD,\ G\text{ and }H\text{ are the mid-points of }CD\text{ and }AC.
\displaystyle \therefore GH=\frac{1}{2}AD\qquad\text{(By the Mid-point Theorem).}
\displaystyle \text{In }\triangle ABC,\ H\text{ and }E\text{ are the mid-points of }AC\text{ and }AB.
\displaystyle \therefore HE=\frac{1}{2}BC\qquad\text{(By the Mid-point Theorem).}
\displaystyle \text{But }AD=BC\qquad\text{(Given).}
\displaystyle \therefore \frac{1}{2}AD=\frac{1}{2}BC.
\displaystyle \therefore EF=FG=GH=HE.
\displaystyle {\therefore EFGH\text{ is a rhombus. Hence proved.}}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{A parallelogram }ABCD\text{ has }P\text{ as the mid-point of }DC\text{ and }Q
\displaystyle \text{as a point on }AC\text{ such that }CQ=\frac{1}{4}AC.\ PQ\text{ produced meets }BC\text{ at }R.
\displaystyle \text{Prove that: }\text{(i) }R\text{ is the mid-point of }BC\qquad\text{(ii) }PR=\frac{1}{2}DB. \displaystyle \textbf{Answer:}
\displaystyle \text{Let the diagonals }AC\text{ and }DB\text{ intersect at }O.
\displaystyle \text{The diagonals of a parallelogram bisect each other.}
\displaystyle \therefore CO=\frac{1}{2}AC.
\displaystyle \text{But }CQ=\frac{1}{4}AC.
\displaystyle \therefore QO=CO-CQ=\frac{1}{2}AC-\frac{1}{4}AC=\frac{1}{4}AC.
\displaystyle \therefore CQ=QO.
\displaystyle \therefore Q\text{ is the mid-point of }CO.
\displaystyle \text{In }\triangle CDO,\ P\text{ and }Q\text{ are the mid-points of }CD\text{ and }CO.
\displaystyle \therefore PQ\parallel DO\qquad\text{(By the Mid-point Theorem).}
\displaystyle \text{Since }D,\ O\text{ and }B\text{ are collinear, }DO\text{ lies on }DB.
\displaystyle \therefore PQ\parallel DB.
\displaystyle \text{Also, }P,\ Q\text{ and }R\text{ are collinear.}
\displaystyle \therefore PR\parallel DB.
\displaystyle \text{In }\triangle DCB,\ P\text{ is the mid-point of }DC\text{ and }PR\parallel DB.
\displaystyle \text{By the converse of the Mid-point Theorem, }R\text{ is the mid-point of }BC.
\displaystyle {\therefore R\text{ is the mid-point of }BC.\text{ Hence proved.}}
\displaystyle \text{Now, }P\text{ and }R\text{ are the mid-points of }DC\text{ and }BC,\text{ respectively.}
\displaystyle \text{Therefore, in }\triangle DCB,\text{ by the Mid-point Theorem,}
\displaystyle PR=\frac{1}{2}DB.
\displaystyle {\therefore PR=\frac{1}{2}DB.\text{ Hence proved.}}
\displaystyle \\

\displaystyle \textbf{Question 10: }D,\ E\text{ and }F\text{ are the mid-points of the sides }AB,\ BC\text{ and }CA
\displaystyle \text{respectively of }\triangle ABC.\ AE\text{ meets }DF\text{ at }O.\ P\text{ and }Q\text{ are the mid-points}
\displaystyle \text{of }OB\text{ and }OC\text{ respectively. Prove that }DPQF\text{ is a parallelogram.}
\displaystyle \textbf{Answer:}
\displaystyle \text{In }\triangle OBC,\ P\text{ and }Q\text{ are the mid-points of }OB\text{ and }OC.
\displaystyle \therefore PQ\parallel BC\qquad\text{(By the Mid-point Theorem).}
\displaystyle \text{In }\triangle ABC,\ D\text{ and }F\text{ are the mid-points of }AB\text{ and }AC.
\displaystyle \therefore DF\parallel BC\qquad\text{(By the Mid-point Theorem).}
\displaystyle \therefore PQ\parallel DF.
\displaystyle \text{In }\triangle AOB,\ D\text{ and }P\text{ are the mid-points of }AB\text{ and }OB.
\displaystyle \therefore DP\parallel AO\qquad\text{(By the Mid-point Theorem).}
\displaystyle \text{In }\triangle AOC,\ F\text{ and }Q\text{ are the mid-points of }AC\text{ and }OC.
\displaystyle \therefore FQ\parallel AO\qquad\text{(By the Mid-point Theorem).}
\displaystyle \therefore DP\parallel FQ.
\displaystyle \text{Thus, both pairs of opposite sides of quadrilateral }DPQF\text{ are parallel}.
\displaystyle {\therefore DPQF\text{ is a parallelogram. Hence proved.}}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{In }\triangle ABC,\ P\text{ is the mid-point of side }BC.\text{ A line through }P\text{ and}
\displaystyle \text{parallel to }CA\text{ meets }AB\text{ at }Q;\text{ and a line through }Q\text{ and parallel to }BC
\displaystyle \text{meets median }AP\text{ at }R.\text{ Prove that: (i) }AP=2AR\quad\text{(ii) }BC=4QR.
\displaystyle \textbf{Answer:} \displaystyle \text{Since }P\text{ is the mid-point of }BC\text{ and }PQ\parallel CA,
\displaystyle \text{by the converse of the Mid-point Theorem, }Q\text{ is the mid-point of }AB.
\displaystyle \text{In }\triangle ABP,\ Q\text{ is the mid-point of }AB.
\displaystyle \text{Also, }QR\parallel BC\text{ and }BP\text{ lies on }BC.
\displaystyle \therefore QR\parallel BP.
\displaystyle \text{By the converse of the Mid-point Theorem, }R\text{ is the mid-point of }AP.
\displaystyle \therefore AR=\frac{1}{2}AP.
\displaystyle {\therefore AP=2AR.\text{ Hence proved.}}
\displaystyle \text{Also, in }\triangle ABP,\ Q\text{ and }R\text{ are the mid-points of }AB\text{ and }AP.
\displaystyle \therefore QR=\frac{1}{2}BP\qquad\text{(By the Mid-point Theorem).}
\displaystyle \text{Since }P\text{ is the mid-point of }BC,\ BP=\frac{1}{2}BC.
\displaystyle \therefore QR=\frac{1}{2}\times\frac{1}{2}BC=\frac{1}{4}BC.
\displaystyle {\therefore BC=4QR.\text{ Hence proved.}}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{In trapezium }ABCD,\ AB\text{ is parallel to }DC;\ P\text{ and }Q\text{ are the}
\displaystyle \text{mid-points of }AD\text{ and }BC,\text{ respectively. }BP\text{ produced meets }DC\text{ produced}
\displaystyle \text{at }E.\text{ Prove that: (i) point }P\text{ bisects }BE\quad\text{(ii) }PQ\text{ is parallel to }AB.
\displaystyle \textbf{Answer:} \displaystyle \text{Since }P\text{ is the mid-point of }AD,\ AP=PD.
\displaystyle \text{Also, }AB\parallel DC\text{ and }D,\ C,\ E\text{ are collinear}.
\displaystyle \therefore AB\parallel DE.
\displaystyle \text{In }\triangle APB\text{ and }\triangle DPE,
\displaystyle AP=PD\qquad\text{(Given),}
\displaystyle \angle APB=\angle DPE\qquad\text{(Vertically opposite angles),}
\displaystyle \angle PAB=\angle PDE\qquad\text{(Alternate angles, since }AB\parallel DE\text{).}
\displaystyle \therefore \triangle APB\cong\triangle DPE\qquad\text{(A.S.A.).}
\displaystyle \therefore BP=PE\qquad\text{(C.P.C.T.C.).}
\displaystyle {\therefore P\text{ bisects }BE.\text{ Hence proved.}}
\displaystyle \text{Now, in }\triangle BEC,\ P\text{ and }Q\text{ are the mid-points of }BE\text{ and }BC.
\displaystyle \therefore PQ\parallel EC\qquad\text{(By the Mid-point Theorem).}
\displaystyle \text{But }EC\text{ lies on }DC\text{ and }DC\parallel AB.
\displaystyle \therefore PQ\parallel AB.
\displaystyle {\therefore PQ\text{ is parallel to }AB.\text{ Hence proved.}}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{In }\triangle ABC,\ AD\text{ is a median and }E\text{ is the mid-point of median }AD.
\displaystyle \text{A line through }B\text{ and }E\text{ meets }AC\text{ at }F.\text{ Prove that }AC=3AF.
\displaystyle \textbf{Answer:} \displaystyle \text{Through }D,\text{ draw }DG\parallel BF,\text{ meeting }AC\text{ at }G.
\displaystyle \text{Since }E\text{ is the mid-point of }AD\text{ and }EF\parallel DG,
\displaystyle \text{by the converse of the Mid-point Theorem, }F\text{ is the mid-point of }AG.
\displaystyle \therefore AF=FG. \qquad\text{... (i)}
\displaystyle \text{Since }AD\text{ is a median of }\triangle ABC,\ D\text{ is the mid-point of }BC.
\displaystyle \text{In }\triangle BCF,\ D\text{ is the mid-point of }BC\text{ and }DG\parallel BF.
\displaystyle \text{By the converse of the Mid-point Theorem, }G\text{ is the mid-point of }CF.
\displaystyle \therefore FG=GC. \qquad\text{... (ii)}
\displaystyle \text{From (i) and (ii), }AF=FG=GC.
\displaystyle AC=AF+FG+GC.
\displaystyle \therefore AC=AF+AF+AF=3AF.
\displaystyle {\therefore AC=3AF.\text{ Hence proved.}}
\displaystyle \\

\displaystyle \textbf{Question 14: }D\text{ and }F\text{ are the mid-points of sides }AB\text{ and }AC\text{ of a triangle}
\displaystyle ABC.\text{ A line through }F\text{ and parallel to }AB\text{ meets }BC\text{ at point }E.
\displaystyle \text{(i) Prove that }BDFE\text{ is a parallelogram.}
\displaystyle \text{(ii) Find }AB,\text{ if }EF=4.8\text{ cm.}
\displaystyle \textbf{Answer:} \displaystyle \text{Since }F\text{ is the mid-point of }AC\text{ and }FE\parallel AB,
\displaystyle \text{by the converse of the Mid-point Theorem, }E\text{ is the mid-point of }BC.
\displaystyle \text{Also, }D\text{ and }F\text{ are the mid-points of }AB\text{ and }AC,\text{ respectively}.
\displaystyle \therefore DF\parallel BC\qquad\text{(By the Mid-point Theorem).}
\displaystyle \text{Since }B,\ E\text{ and }C\text{ are collinear, }DF\parallel BE.
\displaystyle \text{Also, }FE\parallel AB\text{ and }B,\ D\text{ and }A\text{ are collinear}.
\displaystyle \therefore FE\parallel BD.
\displaystyle \text{Thus, both pairs of opposite sides of quadrilateral }BDFE\text{ are parallel}.
\displaystyle {\therefore BDFE\text{ is a parallelogram. Hence proved.}}
\displaystyle \text{Since }F\text{ and }E\text{ are the mid-points of }AC\text{ and }BC,\text{ respectively,}
\displaystyle EF=\frac{1}{2}AB\qquad\text{(By the Mid-point Theorem).}
\displaystyle \therefore 4.8=\frac{1}{2}AB.
\displaystyle \therefore AB=2\times4.8=9.6\text{ cm}.
\displaystyle {\therefore AB=9.6\text{ cm}.}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{In triangle }ABC,\ AD\text{ is the median and }DE,\text{ drawn parallel to side }BA,
\displaystyle \text{meets }AC\text{ at point }E.\text{ Show that }BE\text{ is also a median.}
\displaystyle \textbf{Answer:} \displaystyle \text{Since }AD\text{ is a median of }\triangle ABC,\ D\text{ is the mid-point of }BC.
\displaystyle \therefore BD=DC.
\displaystyle \text{In }\triangle CBA,\ D\text{ is the mid-point of }CB\text{ and }DE\parallel BA.
\displaystyle \text{By the converse of the Mid-point Theorem, }E\text{ is the mid-point of }CA.
\displaystyle \therefore CE=EA.
\displaystyle \text{Thus, }BE\text{ joins vertex }B\text{ to the mid-point }E\text{ of the opposite side }AC.
\displaystyle {\therefore BE\text{ is a median of }\triangle ABC.\text{ Hence proved.}}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{In }\triangle ABC,\ E\text{ is the mid-point of the median }AD\text{ and }BE
\displaystyle \text{produced meets side }AC\text{ at point }Q.\text{ Show that }BE:EQ=3:1.
\displaystyle \textbf{Answer:} \displaystyle \text{Through }D,\text{ draw }DF\parallel BQ,\text{ meeting }AQ\text{ produced at }F.
\displaystyle \text{Since }E\text{ is the mid-point of }AD\text{ and }EB\parallel DF,
\displaystyle \text{by the converse of the Mid-point Theorem, }B\text{ is the mid-point of }AF.
\displaystyle \therefore AB=BF.
\displaystyle \text{Also, }AD\text{ is a median of }\triangle ABC.
\displaystyle \therefore D\text{ is the mid-point of }BC.
\displaystyle \text{In }\triangle BCF,\ D\text{ is the mid-point of }BC\text{ and }DQ\parallel BF.
\displaystyle \text{By the converse of the Mid-point Theorem, }Q\text{ is the mid-point of }CF.
\displaystyle \therefore CQ=QF.
\displaystyle \text{Now, }AF=AB+BF=2AB.
\displaystyle \text{Also, }AQ=AC+CQ=AC+\frac{1}{2}CF.
\displaystyle \text{Since }CF=CQ+QF=2CQ,\text{ we obtain }AQ=\frac{3}{2}AC.
\displaystyle \text{Hence }AQ:QC=3:1.
\displaystyle \text{In }\triangle ADQ,\ E\text{ is the mid-point of }AD\text{ and }BE\parallel DQ.
\displaystyle \text{By the Mid-point Theorem, }B\text{ is the mid-point of }AQ.
\displaystyle \therefore BE:EQ=3:1.
\displaystyle {\therefore BE:EQ=3:1.\text{ Hence proved.}}
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{In the given figure, }M\text{ is the mid-point of }AB\text{ and }DE,\text{ whereas }N
\displaystyle \text{is the mid-point of }BC\text{ and }DF.\text{ Show that }EF=AC. \displaystyle \textbf{Answer:}
\displaystyle \text{In }\triangle ABC,\ M\text{ and }N\text{ are the mid-points of }AB\text{ and }BC.
\displaystyle \therefore MN\parallel AC\text{ and }MN=\frac{1}{2}AC\qquad\text{(By the Mid-point Theorem).}
\displaystyle \text{Also, }M\text{ is the mid-point of }DE\text{ and }N\text{ is the mid-point of }DF.
\displaystyle \text{In }\triangle DEF,\ M\text{ and }N\text{ are the mid-points of }DE\text{ and }DF.
\displaystyle \therefore MN\parallel EF\text{ and }MN=\frac{1}{2}EF\qquad\text{(By the Mid-point Theorem).}
\displaystyle \therefore \frac{1}{2}AC=\frac{1}{2}EF.
\displaystyle \therefore EF=AC.
\displaystyle {\therefore EF=AC.\text{ Hence proved.}}
\displaystyle \\

\displaystyle \textbf{Exercise - 12(B)}

\displaystyle \textbf{Question 1: }\text{Use the following figure to find:}
\displaystyle \text{(i) }BC,\text{ if }AB=7.2\text{ cm.}\qquad\text{(ii) }GE,\text{ if }FE=4\text{ cm.}
\displaystyle \text{(iii) }AE,\text{ if }BD=4.1\text{ cm.}\qquad\text{(iv) }DF,\text{ if }CG=11\text{ cm.} \displaystyle \textbf{Answer:}
\displaystyle \text{The three horizontal lines through }CG,\ BDF\text{ and }AE\text{ are parallel.}
\displaystyle \text{Also, }CD=DE,\text{ as shown in the figure.}
\displaystyle \text{Therefore, by the Equal Intercept Theorem, }AB=BC\text{ and }GF=FE.
\displaystyle \text{(i) }BC=AB=7.2\text{ cm}.
\displaystyle {\therefore BC=7.2\text{ cm}.}
\displaystyle \text{(ii) }GF=FE=4\text{ cm}.
\displaystyle \therefore GE=GF+FE=4+4=8\text{ cm}.
\displaystyle {\therefore GE=8\text{ cm}.}
\displaystyle \text{(iii) Since }AB=BC,\ B\text{ is the mid-point of }AC.
\displaystyle \text{Also, }CD=DE,\text{ so }D\text{ is the mid-point of }CE.
\displaystyle \text{Therefore, in }\triangle ACE,\text{ by the Mid-point Theorem,}
\displaystyle BD=\frac{1}{2}AE.
\displaystyle \therefore AE=2BD=2\times4.1=8.2\text{ cm}.
\displaystyle {\therefore AE=8.2\text{ cm}.}
\displaystyle \text{(iv) In }\triangle CEG,\ D\text{ is the mid-point of }CE\text{ and }DF\parallel CG.
\displaystyle \text{Therefore, by the converse of the Mid-point Theorem, }F\text{ is the mid-point of }EG.
\displaystyle \text{Hence, by the Mid-point Theorem, }DF=\frac{1}{2}CG.
\displaystyle \therefore DF=\frac{1}{2}\times11=5.5\text{ cm}.
\displaystyle {\therefore DF=5.5\text{ cm}.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{In the figure given below, }2AD=AB,\ P\text{ is the mid-point of }AB,
\displaystyle Q\text{ is the mid-point of }DR\text{ and }PR\parallel BS.\text{ Prove that:}
\displaystyle \text{(i) }AQ\parallel BS\qquad\text{(ii) }DS=3RS. \displaystyle \textbf{Answer:}
\displaystyle \text{Since }P\text{ is the mid-point of }AB,\ AP=PB=\frac{1}{2}AB.
\displaystyle \text{But }2AD=AB.
\displaystyle \therefore AD=\frac{1}{2}AB.
\displaystyle \therefore AD=AP.
\displaystyle \therefore A\text{ is the mid-point of }DP.
\displaystyle \text{Also, }Q\text{ is the mid-point of }DR\qquad\text{(Given).}
\displaystyle \text{Therefore, in }\triangle DPR,\ A\text{ and }Q\text{ are the mid-points of }DP\text{ and }DR.
\displaystyle \therefore AQ\parallel PR\qquad\text{(By the Mid-point Theorem).}
\displaystyle \text{But }PR\parallel BS\qquad\text{(Given).}
\displaystyle {\therefore AQ\parallel BS.\text{ Hence proved.}}
\displaystyle \text{Now, }DB=DA+AP+PB.
\displaystyle \text{Since }DA=AP=PB,\ DB=3PB.
\displaystyle \therefore PB=\frac{1}{3}DB.
\displaystyle \therefore DP=DB-PB=DB-\frac{1}{3}DB=\frac{2}{3}DB.
\displaystyle \text{In }\triangle DBS,\ P\text{ lies on }DB,\ R\text{ lies on }DS\text{ and }PR\parallel BS.
\displaystyle \therefore \frac{DR}{DS}=\frac{DP}{DB}=\frac{2}{3}.
\displaystyle \therefore DR=\frac{2}{3}DS.
\displaystyle RS=DS-DR=DS-\frac{2}{3}DS=\frac{1}{3}DS.
\displaystyle {\therefore DS=3RS.\text{ Hence proved.}}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{The side }AC\text{ of }\triangle ABC\text{ is produced to point }E\text{ such that}
\displaystyle CE=\frac{1}{2}AC.\ D\text{ is the mid-point of }BC\text{ and }ED\text{ produced meets }AB\text{ at }F.
\displaystyle \text{Lines through }D\text{ and }C\text{ are drawn parallel to }AB,\text{ meeting }AC\text{ at }P
\displaystyle \text{and }EF\text{ at }R,\text{ respectively. Prove that: (i) }3DF=EF\quad\text{(ii) }4CR=AB.
\displaystyle \textbf{Answer:}
\displaystyle \text{Since }D\text{ is the mid-point of }BC\text{ and }DP\parallel AB,
\displaystyle \text{by the converse of the Mid-point Theorem, }P\text{ is the mid-point of }AC.
\displaystyle \therefore AP=PC=\frac{1}{2}AC.
\displaystyle \text{But }CE=\frac{1}{2}AC\qquad\text{(Given).}
\displaystyle \therefore AP=PC=CE.
\displaystyle \text{The parallel lines }AB,\ PD\text{ and }CR\text{ make equal intercepts }AP\text{ and }PC
\displaystyle \text{on the transversal }ACE.
\displaystyle \therefore FD=DR\qquad\text{(By the Equal Intercept Theorem).}
\displaystyle \text{Also, }PC=CE,\text{ so }C\text{ is the mid-point of }PE.
\displaystyle \text{In }\triangle PDE,\ C\text{ is the mid-point of }PE\text{ and }CR\parallel PD.
\displaystyle \text{By the converse of the Mid-point Theorem, }R\text{ is the mid-point of }DE.
\displaystyle \therefore DR=RE.
\displaystyle \therefore FD=DR=RE.
\displaystyle EF=FD+DR+RE=DF+DF+DF=3DF.
\displaystyle {\therefore 3DF=EF.\text{ Hence proved.}}
\displaystyle \text{Now, in }\triangle PDE,\ C\text{ and }R\text{ are the mid-points of }PE\text{ and }DE.
\displaystyle \therefore CR=\frac{1}{2}PD\qquad\text{(By the Mid-point Theorem).}
\displaystyle \text{Also, in }\triangle ABC,\ P\text{ and }D\text{ are the mid-points of }AC\text{ and }BC.
\displaystyle \therefore PD=\frac{1}{2}AB\qquad\text{(By the Mid-point Theorem).}
\displaystyle \therefore CR=\frac{1}{2}\times\frac{1}{2}AB=\frac{1}{4}AB.
\displaystyle {\therefore 4CR=AB.\text{ Hence proved.}}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{In }\triangle ABC,\text{ the medians }BP\text{ and }CQ\text{ are produced up to points }M
\displaystyle \text{and }N\text{ respectively such that }BP=PM\text{ and }CQ=QN.\text{ Prove that:}
\displaystyle \text{(i) }M,\ A\text{ and }N\text{ are collinear.}\qquad\text{(ii) }A\text{ is the mid-point of }MN.
\displaystyle \textbf{Answer:} \displaystyle \text{Since }BP\text{ is a median of }\triangle ABC,\ P\text{ is the mid-point of }AC.
\displaystyle \therefore AP=PC.
\displaystyle \text{Also, }BP=PM\qquad\text{(Given).}
\displaystyle \therefore P\text{ is the mid-point of }BM.
\displaystyle \text{In }\triangle ABM,\ Q\text{ and }P\text{ are the mid-points of }AB\text{ and }BM.
\displaystyle \therefore QP\parallel AM\text{ and }QP=\frac{1}{2}AM\qquad\text{(By the Mid-point Theorem).}
\displaystyle \text{Since }CQ\text{ is a median of }\triangle ABC,\ Q\text{ is the mid-point of }AB.
\displaystyle \text{Also, }CQ=QN\qquad\text{(Given).}
\displaystyle \therefore Q\text{ is the mid-point of }CN.
\displaystyle \text{In }\triangle ACN,\ P\text{ and }Q\text{ are the mid-points of }AC\text{ and }CN.
\displaystyle \therefore PQ\parallel AN\text{ and }PQ=\frac{1}{2}AN\qquad\text{(By the Mid-point Theorem).}
\displaystyle \therefore AM\parallel PQ\text{ and }AN\parallel PQ.
\displaystyle \therefore AM\text{ and }AN\text{ are the same straight line through }A.
\displaystyle {\therefore M,\ A\text{ and }N\text{ are collinear. Hence proved.}}
\displaystyle \text{Also, }\frac{1}{2}AM=PQ\text{ and }\frac{1}{2}AN=PQ.
\displaystyle \therefore \frac{1}{2}AM=\frac{1}{2}AN.
\displaystyle \therefore AM=AN.
\displaystyle \text{Since }M,\ A\text{ and }N\text{ are collinear and }AM=AN,
\displaystyle {\therefore A\text{ is the mid-point of }MN.\text{ Hence proved.}}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{In }\triangle ABC,\ \angle B\text{ is obtuse. }D\text{ and }E\text{ are the mid-points of}
\displaystyle \text{sides }AB\text{ and }BC\text{ respectively, and }F\text{ is a point on }AC\text{ such that }EF
\displaystyle \text{is parallel to }AB.\text{ Show that }BEFD\text{ is a parallelogram.}
\displaystyle \textbf{Answer:}
\displaystyle \text{Since }E\text{ is the mid-point of }BC\text{ and }EF\parallel BA,
\displaystyle \text{by the converse of the Mid-point Theorem, }F\text{ is the mid-point of }CA.
\displaystyle \text{Also, }D\text{ is the mid-point of }AB.
\displaystyle \text{Therefore, in }\triangle ABC,\ D\text{ and }F\text{ are the mid-points of }AB\text{ and }AC.
\displaystyle \therefore DF\parallel BC\qquad\text{(By the Mid-point Theorem).}
\displaystyle \text{Since }B,\ E\text{ and }C\text{ are collinear, }DF\parallel BE.
\displaystyle \text{Also, }EF\parallel AB\text{ and }B,\ D\text{ and }A\text{ are collinear}.
\displaystyle \therefore EF\parallel BD.
\displaystyle \text{Thus, both pairs of opposite sides of quadrilateral }BEFD\text{ are parallel}.
\displaystyle {\therefore BEFD\text{ is a parallelogram. Hence proved.}}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{In parallelogram }ABCD,\ E\text{ and }F\text{ are the mid-points of the sides}
\displaystyle AB\text{ and }CD,\text{ respectively. The line segments }AF\text{ and }BF\text{ meet }ED\text{ and }EC
\displaystyle \text{at points }G\text{ and }H,\text{ respectively. Prove that:}
\displaystyle \text{(i) }\triangle HEB\cong\triangle FHC\qquad\text{(ii) }GEHF\text{ is a parallelogram.}
\displaystyle \textbf{Answer:} \displaystyle \text{Since }E\text{ is the mid-point of }AB,\ EB=\frac{1}{2}AB.
\displaystyle \text{Since }F\text{ is the mid-point of }CD,\ FC=\frac{1}{2}CD.
\displaystyle \text{But }AB=CD\qquad\text{(Opposite sides of a parallelogram).}
\displaystyle \therefore EB=FC.
\displaystyle \text{Also, }EB\parallel FC,\text{ since }AB\parallel CD.
\displaystyle \therefore EBCF\text{ is a parallelogram}
\displaystyle \text{(one pair of opposite sides is equal and parallel).}
\displaystyle \text{The diagonals of a parallelogram bisect each other.}
\displaystyle \therefore HE=HC\text{ and }HB=HF.
\displaystyle \text{Now, in }\triangle HEB\text{ and }\triangle HFC,
\displaystyle HE=HC\qquad\text{(Proved above),}
\displaystyle HB=HF\qquad\text{(Proved above),}
\displaystyle \angle EHB=\angle CHF\qquad\text{(Vertically opposite angles).}
\displaystyle \therefore \triangle HEB\cong\triangle HFC\qquad\text{(By S.A.S.).}
\displaystyle {\therefore \triangle HEB\cong\triangle FHC.\text{ Hence proved.}}
\displaystyle \text{Also, }AE=\frac{1}{2}AB\text{ and }DF=\frac{1}{2}CD.
\displaystyle \text{Since }AB=CD,\ AE=DF.
\displaystyle \text{Also, }AE\parallel DF,\text{ since }AB\parallel CD.
\displaystyle \therefore AEFD\text{ is a parallelogram}.
\displaystyle \therefore ED\parallel AF\text{ is not required; instead, consider }DEBF.
\displaystyle \text{Since }DF=EB\text{ and }DF\parallel EB,\ DEBF\text{ is a parallelogram}.
\displaystyle \therefore DE\parallel BF.
\displaystyle \text{Since }G,\ E,\ D\text{ are collinear and }H,\ F,\ B\text{ are collinear,}
\displaystyle GE\parallel HF.
\displaystyle \text{Similarly, }AE=FC\text{ and }AE\parallel FC.
\displaystyle \therefore AECF\text{ is a parallelogram}.
\displaystyle \therefore EC\parallel AF.
\displaystyle \text{Since }E,\ H,\ C\text{ are collinear and }A,\ G,\ F\text{ are collinear,}
\displaystyle EH\parallel GF.
\displaystyle \text{Thus, both pairs of opposite sides of }GEHF\text{ are parallel}.
\displaystyle {\therefore GEHF\text{ is a parallelogram. Hence proved.}}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{In }\triangle ABC,\ D\text{ and }E\text{ are points on side }AB\text{ such that}
\displaystyle AD=DE=EB.\text{ Through }D\text{ and }E,\text{ lines are drawn parallel to }BC,\text{ meeting}
\displaystyle AC\text{ at }F\text{ and }G,\text{ respectively. Through }F\text{ and }G,\text{ lines are drawn parallel}
\displaystyle \text{to }AB,\text{ meeting }BC\text{ at }M\text{ and }N,\text{ respectively. Prove that:}
\displaystyle BM=MN=NC.
\displaystyle \textbf{Answer:} \displaystyle \text{Through }A,\text{ draw a line parallel to }BC.
\displaystyle \text{The four parallel lines through }A,\ D,\ E\text{ and }B\text{ cut the transversal }AB
\displaystyle \text{into the equal intercepts }AD,\ DE\text{ and }EB.
\displaystyle \text{Therefore, by the Equal Intercept Theorem, they cut the transversal }AC
\displaystyle \text{into equal intercepts.}
\displaystyle \therefore AF=FG=GC.
\displaystyle \text{Through }C,\text{ draw a line parallel to }AB.
\displaystyle \text{Now, the four parallel lines through }A,\ F,\ G\text{ and }C\text{ cut the}
\displaystyle \text{transversal }AC\text{ into the equal intercepts }AF,\ FG\text{ and }GC.
\displaystyle \text{Therefore, by the Equal Intercept Theorem, they cut the transversal }BC
\displaystyle \text{into equal intercepts }BM,\ MN\text{ and }NC.
\displaystyle {\therefore BM=MN=NC.\text{ Hence proved.}}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{In }\triangle ABC,\ M\text{ is the mid-point of }AB,\ N\text{ is the mid-point}
\displaystyle \text{of }AC\text{ and }D\text{ is any point on the base }BC.\text{ Use the Intercept Theorem to show}
\displaystyle \text{that }MN\text{ bisects }AD.
\displaystyle \textbf{Answer:} \displaystyle \text{Let }MN\text{ intersect }AD\text{ at }P.
\displaystyle \text{Since }M\text{ and }N\text{ are the mid-points of }AB\text{ and }AC,\text{ respectively,}
\displaystyle AM=MB\text{ and }AN=NC.
\displaystyle \text{Thus, the transversals }AB\text{ and }AC\text{ make equal intercepts on the three lines}
\displaystyle \text{through }A,\ MN\text{ and }BC.
\displaystyle \therefore MN\parallel BC\qquad\text{(By the converse of the Equal Intercept Theorem).}
\displaystyle \text{Since }D\text{ lies on }BC,\ MN\parallel BD.
\displaystyle \text{In }\triangle ABD,\ M\text{ is the mid-point of }AB\text{ and }MP\parallel BD.
\displaystyle \text{Therefore, by the converse of the Mid-point Theorem, }P\text{ is the mid-point of }AD.
\displaystyle \therefore AP=PD.
\displaystyle {\therefore MN\text{ bisects }AD.\text{ Hence proved.}}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{If the quadrilateral formed by joining the mid-points of the adjacent}
\displaystyle \text{sides of quadrilateral }ABCD\text{ is a rectangle, show that the diagonals }AC\text{ and }BD
\displaystyle \text{intersect at right angles.}
\displaystyle \textbf{Answer:}
\displaystyle \text{Let }P,\ Q,\ R\text{ and }S\text{ be the mid-points of }AB,\ BC,\ CD\text{ and }DA,
\displaystyle \text{respectively, such that }PQRS\text{ is a rectangle}.
\displaystyle \text{In }\triangle ABC,\ P\text{ and }Q\text{ are the mid-points of }AB\text{ and }BC.
\displaystyle \therefore PQ\parallel AC\qquad\text{(By the Mid-point Theorem).}
\displaystyle \text{In }\triangle BCD,\ Q\text{ and }R\text{ are the mid-points of }BC\text{ and }CD.
\displaystyle \therefore QR\parallel BD\qquad\text{(By the Mid-point Theorem).}
\displaystyle \text{Since }PQRS\text{ is a rectangle, }\angle PQR=90^\circ.
\displaystyle \therefore PQ\perp QR.
\displaystyle \text{But }PQ\parallel AC\text{ and }QR\parallel BD.
\displaystyle \therefore AC\perp BD.
\displaystyle {\therefore \text{The diagonals }AC\text{ and }BD\text{ intersect at right angles. Hence proved.}}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{In }\triangle ABC,\ D\text{ and }E\text{ are the mid-points of the sides }AB
\displaystyle \text{and }AC,\text{ respectively. Through }E,\text{ a straight line is drawn parallel to }AB
\displaystyle \text{to meet }BC\text{ at }F.\text{ Prove that }BDEF\text{ is a parallelogram.}
\displaystyle \text{If }AB=16\text{ cm},\ AC=12\text{ cm and }BC=18\text{ cm, find the perimeter of}
\displaystyle \text{the parallelogram }BDEF.
\displaystyle \textbf{Answer:} \displaystyle \text{Since }E\text{ is the mid-point of }AC\text{ and }EF\parallel AB,
\displaystyle \text{by the converse of the Mid-point Theorem, }F\text{ is the mid-point of }BC.
\displaystyle \text{Also, }D\text{ and }E\text{ are the mid-points of }AB\text{ and }AC,\text{ respectively}.
\displaystyle \therefore DE\parallel BC\qquad\text{(By the Mid-point Theorem).}
\displaystyle \text{Since }B,\ F\text{ and }C\text{ are collinear, }DE\parallel BF.
\displaystyle \text{Also, }EF\parallel AB\text{ and }B,\ D\text{ and }A\text{ are collinear}.
\displaystyle \therefore EF\parallel BD.
\displaystyle \text{Thus, both pairs of opposite sides of quadrilateral }BDEF\text{ are parallel}.
\displaystyle {\therefore BDEF\text{ is a parallelogram. Hence proved.}}
\displaystyle \text{Since }D\text{ is the mid-point of }AB,
\displaystyle BD=\frac{1}{2}AB=\frac{1}{2}\times16=8\text{ cm}.
\displaystyle \text{Since }F\text{ is the mid-point of }BC,
\displaystyle BF=\frac{1}{2}BC=\frac{1}{2}\times18=9\text{ cm}.
\displaystyle \text{Perimeter of parallelogram }BDEF=2(BD+BF).
\displaystyle \therefore \text{Perimeter}=2(8+9)=34\text{ cm}.
\displaystyle {\therefore \text{The perimeter of parallelogram }BDEF\text{ is }34\text{ cm}.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{In the given figure, }AD\text{ and }CE\text{ are medians and }DF\parallel CE.
\displaystyle \text{Prove that }FB=\frac{1}{4}AB. \displaystyle \textbf{Answer:}
\displaystyle \text{Since }CE\text{ is a median of }\triangle ABC,\ E\text{ is the mid-point of }AB.
\displaystyle \therefore BE=\frac{1}{2}AB.
\displaystyle \text{Since }AD\text{ is a median of }\triangle ABC,\ D\text{ is the mid-point of }BC.
\displaystyle \text{In }\triangle BCE,\ D\text{ is the mid-point of }BC\text{ and }DF\parallel CE.
\displaystyle \text{By the converse of the Mid-point Theorem, }F\text{ is the mid-point of }BE.
\displaystyle \therefore FB=\frac{1}{2}BE.
\displaystyle \therefore FB=\frac{1}{2}\times\frac{1}{2}AB=\frac{1}{4}AB.
\displaystyle {\therefore FB=\frac{1}{4}AB.\text{ Hence proved.}}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{In parallelogram }ABCD,\ E\text{ is the mid-point of }AB\text{ and }AP
\displaystyle \text{is parallel to }EC,\text{ which meets }DC\text{ at point }O\text{ and }BC\text{ produced at }P.
\displaystyle \text{Prove that: }\text{(i) }BP=2AD\qquad\text{(ii) }O\text{ is the mid-point of }AP. \displaystyle \textbf{Answer:}
\displaystyle \text{In }\triangle ABP,\ E\text{ is the mid-point of }AB\text{ and }EC\parallel AP.
\displaystyle \text{Therefore, by the converse of the Mid-point Theorem, }C\text{ is the mid-point of }BP.
\displaystyle \therefore BC=CP=\frac{1}{2}BP.
\displaystyle \therefore BP=2BC.
\displaystyle \text{But }BC=AD\qquad\text{(Opposite sides of a parallelogram).}
\displaystyle \therefore BP=2AD.
\displaystyle {\therefore BP=2AD.\text{ Hence proved.}}
\displaystyle \text{Now, }AE\parallel OC,\text{ since }AB\parallel DC.
\displaystyle \text{Also, }AO\parallel EC,\text{ since }O\text{ lies on }AP\text{ and }AP\parallel EC.
\displaystyle \therefore AECO\text{ is a parallelogram}.
\displaystyle \therefore AO=EC\qquad\text{(Opposite sides of a parallelogram).}
\displaystyle \text{Also, in }\triangle ABP,\ E\text{ and }C\text{ are the mid-points of }AB\text{ and }BP.
\displaystyle \therefore EC=\frac{1}{2}AP\qquad\text{(By the Mid-point Theorem).}
\displaystyle \therefore AO=\frac{1}{2}AP.
\displaystyle \text{Since }O\text{ lies on }AP,\ OP=AP-AO.
\displaystyle \therefore OP=AP-\frac{1}{2}AP=\frac{1}{2}AP.
\displaystyle \therefore AO=OP.
\displaystyle {\therefore O\text{ is the mid-point of }AP.\text{ Hence proved.}}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{In trapezium }ABCD,\text{ sides }AB\text{ and }DC\text{ are parallel to}
\displaystyle \text{each other. }E\text{ is the mid-point of }AD\text{ and }F\text{ is the mid-point of }BC.
\displaystyle \text{Prove that }AB+DC=2EF.
\displaystyle \textbf{Answer:} \displaystyle \text{Draw the diagonal }DB\text{ and let it intersect }EF\text{ at }O.
\displaystyle \text{In }\triangle ADB,\ E\text{ is the mid-point of }AD\text{ and }EO\parallel AB.
\displaystyle \text{Therefore, by the converse of the Mid-point Theorem, }O\text{ is the mid-point of }DB.
\displaystyle \therefore EO=\frac{1}{2}AB\qquad\text{(By the Mid-point Theorem).}
\displaystyle \text{In }\triangle DBC,\ O\text{ is the mid-point of }DB\text{ and }OF\parallel DC.
\displaystyle \text{Therefore, by the converse of the Mid-point Theorem, }F\text{ is the mid-point of }BC.
\displaystyle \therefore OF=\frac{1}{2}DC\qquad\text{(By the Mid-point Theorem).}
\displaystyle \text{Since }O\text{ lies on }EF,\ EF=EO+OF.
\displaystyle \therefore EF=\frac{1}{2}AB+\frac{1}{2}DC.
\displaystyle \therefore EF=\frac{1}{2}(AB+DC).
\displaystyle \therefore 2EF=AB+DC.
\displaystyle {\therefore AB+DC=2EF.\text{ Hence proved.}}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{In }\triangle ABC,\ AD\text{ is the median and }DE\text{ is parallel to }BA,
\displaystyle \text{where }E\text{ is a point on }AC.\text{ Prove that }BE\text{ is also a median.}
\displaystyle \textbf{Answer:} \displaystyle \text{Since }AD\text{ is a median of }\triangle ABC,\ D\text{ is the mid-point of }BC.
\displaystyle \therefore BD=DC.
\displaystyle \text{In }\triangle CBA,\ D\text{ is the mid-point of }CB\text{ and }DE\parallel BA.
\displaystyle \text{Therefore, by the converse of the Mid-point Theorem, }E\text{ is the mid-point of }CA.
\displaystyle \therefore CE=EA.
\displaystyle \text{Thus, }BE\text{ joins vertex }B\text{ to the mid-point }E\text{ of the opposite side }AC.
\displaystyle {\therefore BE\text{ is a median of }\triangle ABC.\text{ Hence proved.}}
\displaystyle \\

Discover more from ICSE / ISC / CBSE Mathematics Portal for K12 Students

Subscribe to get the latest posts sent to your email.