Question 1: State True or False:

(i) Two similar polygons are necessarily congruent False

(ii) Two congruent polygons are necessarily similar True

(iii) All equiangular triangles are similar True

(iv) All isosceles triangles are similar False

(v) Two isosceles right angles triangles are similar True

(vi) Two isosceles triangles are similar if an angle of one is congruent to the corresponding angle of the other True

(vii) The diagonals of the trapezium, divide each into proportional segments True

$\displaystyle \\$

Question 2: $\displaystyle \text{In } \triangle ABC, DE \parallel BC$ , where $\displaystyle D \text{ and } E$ are points on $\displaystyle AB \text{ and } AC$ respectively. Prove that $\displaystyle \triangle ADE \sim \triangle ABC .$ Also find the length of $\displaystyle DE, if AD = 12 \text{ cm,} BD = 24 \text{ \text{ cm } } \text{ and } BC = 8 \text{ cm}.$

$\displaystyle \text{In } \triangle ABC, DE \parallel BC$

$\displaystyle \text{Consider } \triangle ABC \text{ and } \triangle ADE$

$\displaystyle \angle ABC = \angle ADE \text{ (alternate angles) }$

$\displaystyle \angle ACB =\angle AED \text{ (alternate angles) }$

$\displaystyle \text{Therefore } \triangle ADE \sim \triangle ABC \text{ (AAA postulate) }$

$\displaystyle \text{Hence } \frac{AD}{AB} = \frac{DE}{BC} = \frac{AE}{AC}$

$\displaystyle \frac{12}{12+24} = \frac{DE}{8}$

$\displaystyle \Rightarrow DE = \frac{12}{36} \times 8 = \frac{8}{3}$

$\displaystyle \\$

Question 3: $\displaystyle \text{Given } \angle GHE = \angle DFE= 90^{\circ}$ , $\displaystyle DH= 8, DF=12, DG=3x-1 \text{ and } DE=4x+2 .$ Find the length of the segments $\displaystyle DG \text{ and } DE.$

$\displaystyle \text{Consider } \triangle EFD \text{ and } \triangle GHD$

$\displaystyle \angle GHE = \angle DFE= 90^{\circ} \text{ (Given) }$

$\displaystyle \angle EDG = \angle HDG (\angle D \text{ is common } )$

$\displaystyle \text{Therefore } \triangle EFD \sim \triangle GHD \text{ (AAA postulate) }$

$\displaystyle \frac{DH}{DF} = \frac{DG}{DE}$

$\displaystyle \frac{8}{12} = \frac{3x-1}{4x+2}$

$\displaystyle 32x+16 = 36x-12$

$\displaystyle 4x=28 \Rightarrow x = 7$

$\displaystyle \text{Substituting } DG = 3 \times 7 -1 = 20$

$\displaystyle DE = 4 \times 7 +2 = 30$

$\displaystyle \\$

Question 4: $\displaystyle D$ is a point of side $\displaystyle BC$ of $\displaystyle \triangle ABC$ such that $\displaystyle \angle ADC = \angle BAC .$ Prove that $\displaystyle CA^2=CB \times CD .$

$\displaystyle \text{Consider } \triangle ABC \text{ and } \triangle ADC$

$\displaystyle \angle BAC = \angle ADC \text{ (Given) }$

$\displaystyle \angle ACB = \angle ACD (\angle C \text{ is common } )$

$\displaystyle \text{Therefore } \triangle ABC \text{ and } \triangle ADC \text{ (AAA postulate) }$

$\displaystyle \text{Therefore } \frac{CA}{CB} = \frac{CD}{CA}$

$\displaystyle \Rightarrow CA^2=CB \times CD$

$\displaystyle \\$

Question 5: In the given figure, $\displaystyle \triangle ABC \text{ and } \triangle AMP$ are right angled at $\displaystyle B \text{ and } M$ respectively. $\displaystyle \text{Given } AC=10 \text{ cm,} AP = 15 \text{ cm } \text{ and } PM= 12 \text{ cm}.$

(i) Prove $\displaystyle \triangle ABC \sim \triangle AMP$           (ii) Find $\displaystyle AB \text{ and } AC$           [2012]

$\displaystyle \text{In } \triangle ABC \text{ and } \triangle AMP$

$\displaystyle \angle BAC = \angle PAM \text{ (common angle) }$

$\displaystyle \angle ABC =\angle PMA$

$\displaystyle \text{Therefore } \triangle ABC \sim \triangle AMP \text{ (AAA postulate) }$

$\displaystyle \text{Since } \triangle ABC \sim \triangle AMP$

$\displaystyle \frac{AP}{AM} = \frac{BC}{PM} = \frac{AC}{AP}$

$\displaystyle \text{Given } AC=10 \text{ cm,} AP = 15 \text{ cm } \text{ and } PM= 12 \text{ cm }$

$\displaystyle \Rightarrow AB = \frac{10}{15} \times 11 = 7.33 \text{ cm }$

$\displaystyle \Rightarrow BC = \frac{10}{15} \times 12 = 8 \text{ cm }$

$\displaystyle \\$

Question 6: $\displaystyle E \text{ and } F$ are points in sides $\displaystyle DC \text{ and } AB$ respectively of parallelogram $\displaystyle ABCD .$ If diagonal $\displaystyle AC$ and segment $\displaystyle EF$ intersect at $\displaystyle G$ ; prove that: $\displaystyle AG \times EG = FG \times CG .$

$\displaystyle \text{In } \triangle AGF \text{ and } \triangle EGC$

$\displaystyle \angle AGF = \angle EGC \text{ (opposite angles) }$

$\displaystyle \angle ECG =\angle FAG \text{ (alternate angles) }$

$\displaystyle \text{Therefore } \triangle AGF \sim \triangle ECG \text{ (AAA postulate) }$

$\displaystyle \text{Therefore } \frac{EG}{FG} = \frac{CG}{AG}$

$\displaystyle \Rightarrow AG \times EG = FG \times CG$

$\displaystyle \\$

Question 7: $\displaystyle \text{Given } RS \text{ and } PT$ are altitudes of $\displaystyle \triangle PQR .$ Prove that $\displaystyle \text{(i) } \triangle PQT \sim \triangle QRS \hspace{1.0cm} \text{(ii) } PQ \times QS = RQ \times QT$

$\displaystyle \text{(i) } \text{Consider } \triangle PQT \text{ and } \triangle QRS$

$\displaystyle \angle QTP = \angle QSR= 90^{\circ}$ (Given, altitudes)

$\displaystyle \angle PQT = \angle SQR \ \ \ (\angle Q \text{ is common } )$

$\displaystyle \text{Therefore } \triangle PQT \sim \triangle QRS \text{ (AAA postulate) }$

$\displaystyle \text{(ii) } \frac{PQ}{RQ} = \frac{QT}{QS}$

$\displaystyle PQ \times QS = RQ \times QT$

$\displaystyle \\$

Question 8: $\displaystyle \text{Given } ABCD$ is a rhombus, $\displaystyle DPR \text{ and } CBR$ are straight lines. Prove that $\displaystyle DP \times CR = DC \times PR .$

$\displaystyle \text{In } \triangle DPA \text{ and } \triangle RPC$

$\displaystyle \angle DPA = \angle RPC \text{ (vertically opposite angles) }$

$\displaystyle \angle PAD =\angle PCR \text{ (alternate angles) }$

$\displaystyle \text{Therefore } \triangle DPA \sim \triangle RPC \text{ (AAA postulate) }$

$\displaystyle \text{Therefore } \frac{DB}{PR} = \frac{AD}{CR}$

$\displaystyle \frac{DP}{PR} = \frac{DC}{CR} \ \ \ ( AD = DC \text{ as } ABCD \text{ is a Rhombus} )$

$\displaystyle \Rightarrow DP \times CR = DC \times PR$

$\displaystyle \\$

$\displaystyle \text{Question 9: Given } FB=FD, AE \perp FD \text{ and } FC \perp AD . \\ \\ \text{Prove } \frac{GB}{AD} = \frac{BC}{ED}$

$\displaystyle FB=FD, AE \perp FD \text{ and } FC \perp AD .$

$\displaystyle \text{Therefore } \angle FDB=\angle FBD$

$\displaystyle \text{Consider } \triangle AED \text{ and } \triangle FCB$

$\displaystyle \angle AED = \angle FCB= 90^{\circ} \text{ (Given) }$

$\displaystyle \angle ADE = \angle FBC$

$\displaystyle \text{Therefore } \triangle AED \sim \triangle FBC \text{ (AAA postulate) }$

$\displaystyle \frac{AD}{FB} = \frac{ED}{BC}$

$\displaystyle \frac{FB}{AD} = \frac{BC}{ED}$

$\displaystyle \\$

Question 10: $\displaystyle \text{In } \triangle ABC, \angle B = 2 \angle C$ and the bisector of $\displaystyle \angle B$ meets $\displaystyle CA$ at point $\displaystyle D .$ Prove that $\displaystyle \text{(i) } \triangle ABC \sim \triangle ABD \hspace{1.0cm} \text{(ii) } DC:AD = BC:AB$

$\displaystyle \text{Given } \triangle ABC, \angle B = 2 \angle C$

$\displaystyle \angle ABD = \angle DBC$

$\displaystyle \text{Therefore } \angle ABD = \angle DBC = \angle ACB$

$\displaystyle \text{Consider } \triangle ABC \text{ and } \triangle ABD$

$\displaystyle \angle BAC = \angle DAB \text{ (Given) }$

$\displaystyle \angle ACB = \angle ABD$

$\displaystyle \text{Therefore } \triangle ABC \sim \triangle ABD \text{ (AAA postulate) }$

$\displaystyle \frac{BC}{BD} \frac{AB}{AD}$

$\displaystyle \frac{BC}{AB} = \frac{BD}{AD}$

$\displaystyle \frac{BC}{AB} = \frac{DC}{AD}$ ( $\displaystyle \angle DBC = \angle DCB \Rightarrow DC = BD$ )

$\displaystyle DC:AD = BC:AB$

$\displaystyle \\$

Question 11: $\displaystyle \text{In } \triangle PQR, \angle Q=90^{\circ} \text{ and } QM \perp PR .$ Prove that:

$\displaystyle \text{(i) } PQ^2 = PM \times PR$

$\displaystyle \text{(ii) } QR^2 = PR \times MR$

$\displaystyle \text{(iii) } PQ^2 + QR^2 = PR^2$

$\displaystyle \text{(i) } \text{Given } \triangle PQR, \angle Q=90^{\circ} \text{ and } QM \perp PR$

$\displaystyle \text{Consider } \triangle PQM \text{ and } \triangle PRQ$

$\displaystyle \angle PMQ = \angle PQR = 90^{\circ} \text{ (Given) }$

$\displaystyle \angle QPM =\angle RPQ \text{ (common angle) }$

$\displaystyle \text{Therefore } \triangle PQM \sim \triangle PRQ \text{ (AAA postulate) }$

$\displaystyle \text{Hence } \frac{PQ}{PR} = \frac{PM}{PQ}$

$\displaystyle \text{Therefore } PQ^2 = PM \times PR$

$\displaystyle \text{(ii) } \text{Consider } \triangle QMR \text{ and } \triangle PQR$

$\displaystyle \angle QMR = \angle PQR = 90^{\circ} \text{ (Given) }$

$\displaystyle \angle QRM =\angle QRP \text{ (common angle) }$

$\displaystyle \text{Therefore } \triangle QMR \sim \triangle PQR \text{ (AAA postulate) }$

$\displaystyle \text{Hence } \frac{QR}{PR} = \frac{MR}{QR}$

$\displaystyle \text{Therefore } QR^2 = PR \times MR$

$\displaystyle PQ^2 + QR^2 = PM \times PR + PR \times MR$

$\displaystyle \Rightarrow PQ^2 + QR^2 = PR \times (PM + MR)$

$\displaystyle \Rightarrow PQ^2 + QR^2 = PR \times PR = PR^2$

$\displaystyle \\$

Question 12: $\displaystyle \text{In } \triangle ABC, \angle C= 90^{\circ}, CD \perp AB .$ Prove that $\displaystyle CD^2 = AD \times DB .$

$\displaystyle \triangle ABC, \angle C= 90^{\circ}, CD \perp AB .$

$\displaystyle \text{Consider } \triangle PQM \text{ and } \triangle PRQ$

$\displaystyle \angle PMQ = \angle PQR = 90^{\circ} \text{ (Given) }$

$\displaystyle \angle QPM =\angle RPQ \text{ (common angle) }$

$\displaystyle \text{Therefore } \triangle PQM \sim \triangle PRQ \text{ (AAA postulate) }$

$\displaystyle \text{Hence } \frac{PQ}{PR} = \frac{PM}{PQ}$

$\displaystyle \text{Therefore } PQ^2 = PM \times PR$

$\displaystyle \\$

Question 13: $\displaystyle \text{In } \triangle ABC, \angle B= 90^{\circ}, BD \perp AC .$

$\displaystyle \text{(i) If } CD=10 \text{ cm } \text{ and } BD = 8 \text{ cm, Find } AD$

$\displaystyle \text{(ii) If } AC=18 \text{ cm } \text{ and } AD = 6 \text{ cm, Find } BD$

$\displaystyle \text{(iii) If } AC=9 \text{ cm } \text{ and } AB = 7 \text{ cm, Find } AD$

$\displaystyle \triangle ABC, \angle B= 90^{\circ}, BD \perp AC$

$\displaystyle \text{Consider } \triangle CDB \text{ and } \triangle BCD$

$\displaystyle \angle ADB = \angle BDC = 90^{\circ} \text{ (Given) }$

$\displaystyle \angle ABD =\angle BCD$

$\displaystyle \text{Therefore } \triangle CDB \sim \triangle BDA \text{ (AAA postulate) }$

$\displaystyle \text{(i) } \frac{CD}{BD} = \frac{BD}{AD}$

$\displaystyle \text{Therefore } AD = \frac{BD^2}{CD} = \frac{8^2}{10} = 6.4 \text{ cm }$

$\displaystyle \text{(ii) } \frac{BD}{DA} = \frac{CD}{BD}$

$\displaystyle \text{Therefore } BD = \sqrt{CD \times DA} = \sqrt{72}= 8.49 \text{ cm }$

$\displaystyle \text{(iii) } \frac{AD}{AB} = \frac{AB}{AC}$

$\displaystyle \text{Therefore } AD = \frac{AB \times AB}{AC} = \frac{7 \times 7}{9} = 5.44 \text{ cm }$

$\displaystyle \\$

Question 14: In the figure, $\displaystyle PQRS$ is a parallelogram with $\displaystyle PQ=16 \text{ cm } \text{ and } QR=10 \text{ cm } .$ $\displaystyle L$ is a point on $\displaystyle PR$ such that $\displaystyle RL:LP=2:3 .$

$\displaystyle QL$ produced meets $\displaystyle RS$ at $\displaystyle M \text{ and } PS$ produced at $\displaystyle N .$ Find the lengths of $\displaystyle PN \text{ and } RM .$ [1997]

$\displaystyle \text{In } \triangle RLQ \text{ and } \triangle PLN$

$\displaystyle \angle RLQ = \angle PLN \text{ (vertically opposite angles) }$

$\displaystyle \angle LRQ = \angle LPN \text{ (alternate angles) }$

$\displaystyle \text{Therefore } \triangle RLQ \sim \triangle PLN \text{ (AAA postulate) }$

$\displaystyle \text{Therefore } \frac{RL}{LP} = \frac{RQ}{PN}$

$\displaystyle \Rightarrow PN = \frac{10}{2} \times 3 = 15 \text{ cm }$

$\displaystyle \text{In } \triangle RLM \text{ and } \triangle PLQ$

$\displaystyle \angle RLM = \angle PLQ \text{ (vertically opposite angles) }$

$\displaystyle \angle LRM = \angle LPQ \text{ (alternate angles) }$

$\displaystyle \text{Therefore } \triangle RLM \sim \triangle PLQ \text{ (AAA postulate) }$

$\displaystyle \text{Therefore } \frac{RM}{PQ} = \frac{RL}{LP}$

$\displaystyle \Rightarrow RM = \frac{2}{3} \times 16 = 10.667 \text{ cm }$

$\displaystyle \\$

Question 15: In quadrilateral $\displaystyle ABCD$ , Diagonal $\displaystyle AC \text{ and } BD$ intersect at point $\displaystyle E$ such that: $\displaystyle AE:EC=BE:ED .$ Show that $\displaystyle ABCD$ is a parallelogram.

$\displaystyle \text{Given } \frac{AE}{EC} = \frac{BE}{ED} .$.. … … … (i)

$\displaystyle \text{In } \triangle ADB \text{, because } EF \parallel AB \text{, by proportionality theorem }$

$\displaystyle \frac{DF}{FA} = \frac{ED}{BE} .$.. … … … (i)

Therefore from (i) and (ii)

$\displaystyle \frac{DF}{FA} = \frac{EC}{AE}$

Similarly $\displaystyle \text{In } \triangle DCA$

$\displaystyle \frac{DF}{FA} = \frac{CE}{AE} \Rightarrow FE \parallel DC$

Also it is given that $\displaystyle FE \parallel AB$

$\displaystyle \text{Therefore } DC \parallel AB .$ $\displaystyle \text{Therefore } ABCD \text{ is a parallelogram. }$

$\displaystyle \\$

Question 16: $\displaystyle \text{Given } AB \parallel DE \text{ and } BC \parallel EF .$ Prove that:

$\displaystyle \text{(i) } \frac{AD}{DG} = \frac{CF}{FG}$

$\displaystyle \text{(ii) } \triangle DFG \sim \triangle ACG$

$\displaystyle \text{(i) } \text{Given } AB \parallel DE \text{ and } BC \parallel EF$

$\displaystyle \text{In } \triangle AGB, \frac{AD}{DG} = \frac{EB}{GE} .$.. … … … (i)

$\displaystyle \text{In } \triangle BGC, \frac{EB}{GE} = \frac{CF}{FG} .$.. … … … (ii)

$\displaystyle \text{From (i) and (ii) you get } \frac{AD}{DG} = \frac{CF}{FG}$

$\displaystyle \text{(ii) } \text{Consider } \triangle DFG \text{ and } \triangle ACG$

$\displaystyle \text{We have } \frac{AD}{DG} = \frac{CF}{FG}$

and $\displaystyle \angle DGF = \angle AGC \text{ (common angle) }$

$\displaystyle \text{Therefore } \triangle DFG \sim \triangle ACG$

$\displaystyle \\$

Question 17: $\displaystyle \text{In } \triangle ABC, AD \perp BC \text{ and } AD^2 = BD \times DC .$ Show that $\displaystyle \angle BAC = 90^{\circ} .$

$\displaystyle \text{Given } AD^2 = BD \times DC$

$\displaystyle \Rightarrow \frac{AD}{DC} = \frac{BD}{AD}$

Also $\displaystyle \text{Given } \angle ADB = \angle ADC = 90^{\circ}$

$\displaystyle \text{Therefore } \triangle DBA \sim \triangle DAC$ (SAS postulate)

$\displaystyle \text{Therefore } \angle C = \angle BAD$

$\displaystyle \angle B = \angle DAC$

Adding $\displaystyle \angle C + \angle B = \angle BAD + \angle DAC$

$\displaystyle \Rightarrow \angle A =\angle C + \angle B .$.. … … (i)

We k$\displaystyle \text{Now } \angle A + \angle B + \angle C = 180^{\circ} .$.. … … … (ii)

From (i) and $\displaystyle \text{(ii) } 2 \angle A = 180^{\circ}$

$\displaystyle \Rightarrow \angle A = 90^{\circ}$

$\displaystyle \\$

Question 18: In the given figure $\displaystyle AB \parallel EF \parallel DC$ ; $\displaystyle AB=67.5 \text{ cm,} DC = 40.5 \text{ cm } \text{ and } AE = 52.5 \text{ cm } .$

(i) Name the three pairs of similar triangles

(ii) Find the lengths of $\displaystyle EC \text{ and } EF$

$\displaystyle \text{Given } AB \parallel EF \parallel DC$ , The three pairs of similar triangles are

$\displaystyle \triangle ABC \sim \triangle CEF$

$\displaystyle \triangle BCD \sim \triangle BEF$

$\displaystyle \triangle ABE \sim \triangle CDE$

$\displaystyle \text{Since } \triangle ABE \sim \triangle CDE$

$\displaystyle \frac{AB}{CD} = \frac{AE}{CE} \Rightarrow CE = \frac{52.5 \times 40.5}{67.5} = 31.5 \text{ cm }$

$\displaystyle \text{Since } \triangle ABC \sim \triangle CEF$

$\displaystyle \frac{CE}{CA} = \frac{EF}{AB} \Rightarrow EF = \frac{67.5 \times 31.5}{52.5+31.5} = 25.3125 \text{ cm }$

$\displaystyle \\$

$\displaystyle \text{Question 19: In the given figure } QR \parallel AB \text{, and } DR \parallel QB. \\ \text{ Prove that } PQ^2 = PD \times PA.$

$\displaystyle \text{Given } QR \parallel AB \text{, and } DR \parallel QB$

Using basic proportionality theorem

$\displaystyle \text{In } \triangle PQR \text{ and } \triangle PAB$

$\displaystyle \frac{PQ}{PA} = \frac{PR}{PB} .$.. … … … (i)

$\displaystyle \text{In } \triangle PDR \text{ and } \triangle PQB$

$\displaystyle \frac{PD}{PQ} = \frac{PR}{PB} .$.. … … … (ii)

From (i) and (ii) we get

$\displaystyle \frac{PQ}{PA} = \frac{PD}{PQ}$

$\displaystyle \Rightarrow PQ^2 = PD \times PA$

$\displaystyle \\$

Question 20: Through the mid point $\displaystyle M$ of the side $\displaystyle CD$ of a parallelogram $\displaystyle ABCD$ , the line $\displaystyle BM$ is drawn intersecting diagonal $\displaystyle AC \text{ in } L \text{ and } AD$ produced $\displaystyle \text{In } E .$ Prove that $\displaystyle EL=2BL .$

$\displaystyle \text{Consider } \triangle DEM \text{ and } \triangle CBM$

$\displaystyle \angle AEB = \angle EBC \text{ (alternate angles) }$

$\displaystyle \angle EMD = \angle CMB \text{ (opposite angles) }$

$\displaystyle DM = MC$ (as M is the midpoint of CD)

$\displaystyle \text{Therefore } \triangle DEM \cong \triangle CBM$

$\displaystyle \text{Therefore } DE = BC \text{ (corresponding angles) }$

$\displaystyle AD = BC \text{ (opposite sides of a parallelogram) }$

$\displaystyle \text{Therefore } AE = AD + DE = 2BC$

$\displaystyle \text{Now } \text{Consider } \triangle ELA \text{ and } \triangle LBC$

$\displaystyle \angle AEB = \angle EBC$

$\displaystyle \angle ELA = \angle EBC$

$\displaystyle \text{Therefore } \triangle ELA \sim \triangle LBC$

$\displaystyle \text{Hence } \frac{EL}{BL} = \frac{EA}{BC}$

$\displaystyle \frac{EL}{BL} = \frac{2BC}{BC} = 2$

$\displaystyle \text{Hence } EL = 2 BL$

$\displaystyle \\$

Question 21: In the figure given below, $\displaystyle P$ is a point on $\displaystyle AB$ such that $\displaystyle AP:PB=4:3 .$ $\displaystyle PQ \parallel AC .$

(i) Calculate the ratio $\displaystyle PQ:AC$ , giving reasons for your answer.

$\displaystyle \text{(ii) } \text{In } \triangle ARC, \angle ARC = 90^{\circ} \text{ and } \text{In } \triangle PQS, \angle PSQ=90^{\circ} .$ $\displaystyle \text{Given } QS=6 \text{ cm }$ , calculate length of $\displaystyle AR .$ [1999]

$\displaystyle \text{(i) } \text{Given } AP:PB=4:3$

Also $\displaystyle PQ \parallel AC$ , applying basic proportionality theorem

$\displaystyle \frac{AP}{PB} = \frac{CQ}{QB}$

$\displaystyle \Rightarrow \frac{CQ}{QB} = \frac{4}{3}$

$\displaystyle \Rightarrow \frac{BQ}{BC} = \frac{3}{7}$

$\displaystyle \angle QBP = \angle ACB \text{ (corresponding angles) }$

$\displaystyle \angle QPB = \angle CAB \text{ (corresponding angles) }$

$\displaystyle \text{Therefore } \triangle PBQ \sim \triangle ABC \text{ (AAA postulate) }$

$\displaystyle \frac{PQ}{ AC} = \frac{BQ}{ BC} = \frac{3}{ 7}$

$\displaystyle \text{(ii) } \text{Given } \angle ARC = \angle QSP = 90^{\circ}$

$\displaystyle \angle ACR = \angle SPQ \text{ (alternate angles) }$

$\displaystyle \text{Therefore } \triangle ARC \sim \triangle QSP \text{ (AAA postulate) }$

$\displaystyle \frac{AR}{ SQ} = \frac{AC}{ PQ} = \frac{7}{ 3}$

$\displaystyle \Rightarrow AR = \frac{7 \times 6}{3} = 14 \text{ cm }$

$\displaystyle \\$

Question 22: In the right angled $\displaystyle \triangle QPR, PM$ is the altitude. Given that $\displaystyle QR= 8 \text{ cm } \text{ and } MQ = 3.5 \text{ cm }$ , calculate the value of $\displaystyle PR .$ [2000]

$\displaystyle \angle QPR = \angle PMR = 90^{\circ}$

$\displaystyle \angle PRQ = \angle PRM \text{ (common) }$

$\displaystyle \triangle PQR \sim \triangle MPR \text{ (AAA postulate) }$

$\displaystyle \frac{QR}{ PR} = \frac{ PR}{ MR}$

$\displaystyle PR^2 = QR \times MR = 8 \times 4.5 = 36 \Rightarrow PR = 6 \text{ cm }$

$\displaystyle \\$

Question 23: In the figure given below, median $\displaystyle BD \text{ and } CE$ of the $\displaystyle \triangle ABC$ meet at $\displaystyle G .$ Prove that:

$\displaystyle \text{(i) } \triangle EGD \sim \triangle CGB$

$\displaystyle \text{(ii) } BG = 2 GD \text{ from (i)}$

$\displaystyle \text{Given } BD \text{ and } CE$ are medians

$\displaystyle \text{Therefore } AD = DC \text{ and } AE = BE$

Applying converse of proportionality theorem $\displaystyle ED \parallel BC$

$\displaystyle \text{In } \triangle EGD \text{ and } \triangle CGB$

$\displaystyle \angle DEG = \angle GCB \text{ (alternate angles) }$

$\displaystyle \angle EGD = \angle BGC \text{ (vertically opposite angles) }$

$\displaystyle \text{Therefore } \triangle EGD \sim \triangle CGB \text{ (AAA postulate) }$

$\displaystyle \text{Therefore } \frac{GD}{GB} = \frac{ED}{BC}$

$\displaystyle \text{In } \triangle EAD \text{ and } \triangle BAC$

$\displaystyle \angle AED = \angle ABC \text{ (corresponding angles) }$

$\displaystyle \angle EAD = \angle BAC \text{ (common angle) }$

$\displaystyle \text{Therefore } \triangle EAD \sim \triangle BAC \text{ (AAA postulate) }$

$\displaystyle \text{Therefore } \frac{ED}{BC} = \frac{AE}{AB} = \frac{1}{2}$

$\displaystyle \frac{ED}{BC} = \frac{1}{2}$

$\displaystyle \text{From 1, } \frac{GD}{GB} = \frac{1}{2}$

$\displaystyle GB = 2GD$