\displaystyle \textbf{Chapter: Mid-Point Theorem (Proof and simple applications and its converse)}

\displaystyle \textbf{Theorem 6}
\displaystyle \textit{\text{The line segment joining the mid-points of any two sides of a triangle is}}
\displaystyle \textit{\text{parallel to the third side, and is equal to half of it.}}
\displaystyle \textbf{Given: }\text{D and E are the mid-points of sides }AB\text{ and }AC
\displaystyle \text{respectively of }\triangle ABC.
\displaystyle \textbf{To Prove: }DE\parallel BC\text{ and }DE=\frac{1}{2}BC.
\displaystyle \textbf{Construction: Draw }CF\text{ parallel to }BA\text{ which meets }DE\displaystyle \text{produced at }F.
\displaystyle \textbf{Proof:}
\displaystyle \textbf{Statement:}\hspace{2.2cm}\textbf{Reason:}
\displaystyle \text{1. In }\triangle ADE\text{ and }\triangle CEF:
\displaystyle \text{(i) }AE=EC\hspace{2.5cm}\text{Given, E is the mid-point of }AC.
\displaystyle \text{(ii) }\angle AED=\angle CEF\hspace{1.2cm}\text{Vertically opposite angles.}
\displaystyle \text{(iii) }\angle EAD=\angle ECF\hspace{1.1cm}\text{Alternate angles.}
\displaystyle \therefore\ \triangle ADE\cong\triangle CFE\hspace{0.6cm}\text{A.S.A.}
\displaystyle \text{2. }\therefore\ AD=CF\hspace{2.5cm}\text{Corresponding parts of congruent triangles.}
\displaystyle \text{3. But, }AD=BD\hspace{2.3cm}\text{Given, D is the mid-point of }AB.
\displaystyle \text{4. }\therefore\ CF=BD\hspace{2.3cm}\text{From 2 and 3.}
\displaystyle \text{5. }\therefore\ BCFD\text{ is a parallelogram}\hspace{0.4cm}\text{Opposite sides }CF\text{ and }BD
\displaystyle \hspace{6.3cm}\text{are equal and parallel.}
\displaystyle \therefore\ DF\parallel BC\text{ and so, }DE\parallel BC.\hspace{0.5cm}\text{Opposite sides of a parallelogram are parallel.}
\displaystyle \hspace{5.0cm}\textbf{(First part proved)}
\displaystyle \text{Now, }DE=EF\hspace{2.8cm}\triangle ADE\cong\triangle CFE
\displaystyle \hspace{1.9cm}=\frac{1}{2}DF
\displaystyle \hspace{1.9cm}=\frac{1}{2}BC\hspace{2.0cm}\text{In parallelogram }BCFD;\ DF=BC.
\displaystyle \therefore\ DE\parallel BC\text{ and }DE=\frac{1}{2}BC.
\displaystyle \therefore\ \textbf{Hence Proved.}

\displaystyle \textbf{Theorem 7 (Converse of Mid-point Theorem)}
\displaystyle \textit{\text{The straight line drawn through the mid-point of one side of a triangle parallel to another,}}
\displaystyle \textit{\text{bisects the third side.}}
\displaystyle \textbf{Given: }\text{D is the mid-point of side }AB\text{ of }\triangle ABC\text{ and }DE
\displaystyle \text{is drawn parallel to the side }BC.
\displaystyle \textbf{To Prove: }DE\text{ bisects }AC,\text{ i.e. }AE=EC.
\displaystyle \textbf{Construction: Draw }CF\text{ parallel to }BA\text{ which meets }DE\displaystyle \text{produced at }F.
\displaystyle \textbf{Proof:}
\displaystyle \textbf{Statement:}\hspace{2.2cm}\textbf{Reason:}
\displaystyle \text{1. }BCFD\text{ is a parallelogram}\hspace{0.7cm}DF\parallel BC\text{ and }CF\parallel BD.
\displaystyle \text{2. }CF=BD\hspace{2.8cm}\text{Opposite sides of a parallelogram are equal.}
\displaystyle \text{3. }CF=DA\hspace{2.8cm}\text{Since, }BD=DA\text{ (given).}
\displaystyle \text{4. In }\triangle ADE\text{ and }\triangle CFE:
\displaystyle \text{(i) }AD=CF\hspace{2.5cm}\text{From (3).}
\displaystyle \text{(ii) }\angle DAE=\angle ECF\hspace{1.2cm}\text{Alternate angles.}
\displaystyle \text{(iii) }\angle ADE=\angle EFC\hspace{1.2cm}\text{Alternate angles.}
\displaystyle \therefore\ \triangle ADE\cong\triangle CFE\hspace{0.6cm}\text{A.S.A.}
\displaystyle \therefore\ AE=EC\hspace{2.9cm}\text{Corresponding parts of congruent triangles are equal.}
\displaystyle \therefore\ \textbf{Hence Proved.}

\displaystyle \\

\displaystyle \textbf{Note: }\text{In the figure given above, D is the mid-point of }AB\text{ and E is proved}
\displaystyle \text{to be the mid-point of }AC,\text{ therefore, }DE\text{ will be half of the third side,}
\displaystyle \text{i.e., }DE=\frac{1}{2}BC.

\displaystyle \textbf{Equal Intercept Theorem (Proof and simple application)}
\displaystyle \textbf{Theorem 8}
\displaystyle \textit{\text{If a transversal makes equal intercepts on three or more parallel lines, then any other line}}
\displaystyle \textit{\text{cutting them will also make equal intercepts.}}
\displaystyle \textbf{Given: }\text{Transversal } \\ AB\text{ makes equal intercepts on three parallel lines}
\displaystyle l\parallel m\parallel n\text{ and } \\ PQ=QR.
\displaystyle \text{CD is another transversal which} \\ \text{makes intercepts }LM\text{ and }MN.
\displaystyle \textbf{To Prove: }LM=MN.
\displaystyle \textbf{Construction: Draw }PS \\ \text{ and }QT\text{ parallel to }CD.
\displaystyle \textbf{Proof:}
\displaystyle \textbf{Statement:}\hspace{2.2cm}\textbf{Reason:}
\displaystyle \text{1. In }\triangle PQS \\ \text{ and }\triangle QRT:
\displaystyle \text{(i) }PQ=QR\hspace{2.7cm}\text{Given.}
\displaystyle \text{(ii) }\angle PQS=\angle QRT \\ \hspace{1.1cm}\text{Corresponding angles.}
\displaystyle \text{(iii) }\angle QPS=\angle RQT\hspace{1.1cm} \\ \text{Corresponding angles as }PS\parallel CD\parallel QT.
\displaystyle \therefore\ \triangle PQS\cong\triangle QRT\hspace{0.5cm}\text{A.S.A.}
\displaystyle \text{2. }\therefore\ PS=QT\hspace{2.5cm}\text{Corresponding parts of congruent triangles are equal.}
\displaystyle \text{3. }PSML\text{ is a parallelogram}\hspace{0.5cm}\text{Both the pairs of opposite sides are parallel.}
\displaystyle \therefore\ PS=LM\hspace{2.4cm}\text{Opposite sides of a parallelogram are equal.}
\displaystyle \text{4. }QTNM\text{ is a parallelogram}\hspace{0.4cm}\text{Both the pairs of opposite sides are parallel.}
\displaystyle \therefore\ QT=MN\hspace{2.4cm}\text{Opposite sides of a parallelogram are equal.}
\displaystyle \text{5. }\therefore\ LM=MN\hspace{2.3cm}\text{From (2), (3) and (4).}
\displaystyle \therefore\ \textbf{Hence Proved.}

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