SChand Composite Mathematics Class 7 Chapter 13 Congruence of Triangle Exercise 13B

 Exercise 13 B

Question 1

AB and CD bisect each other at K. Prove that AC = BD .

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Sol: AK= BK

$\angle A K C=\angle B K D$ (Vertically opposite angle )

CK = DK 

By SAS $\triangle A K C, \cong \triangle B K D$

C.P.C.T $\quad A C=B D$ Hence proved 

Question 2

The sides BA and CA have been produced such that BA = AD and CA = AE prove that DE||BC 

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Sol: $A B=A D$

$\angle B A C=\angle E A D$  (Vertically opposite angle )

by $S A S \quad \triangle A B C \cong \triangle A E D$

by C.P.CT $\angle B=\angle D \quad \because B C \| D E$  Hence proved .

Question 3

$O A=O B, O C=O D$, $\angle A O B=\angle C O D$ prove that AC = BD 

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Sol: 

$\begin{aligned} O A &=O B \\ \angle A O B &=\angle C O D \\ \angle A O C+\angle C O B=\angle C O B+\angle B O D \\ \angle A O C &=\angle B O D \\ O C &=O D \end{aligned}$

By SAS $\triangle A O C \cong \triangle B O D$

by C.P.CT $A C=B D$ hence proved 

Question 4

If AB and MN bisect each other at O and AM and BN are $\perp$ on xy. Prove that $\triangle S$, OAM and OBN are congruent and hence prove that AM= BN 

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Sol: $\angle A M O=\angle B N O=90^{\circ}$

$O B=O A$

$O M=O N$

by RHS $\triangle A M O \cong \triangle B N O$ 

by $C \cdot P \cdot c \cdot T \quad A M=B N$

Question 5

$\angle XYZ$ is bisected by YP . L is any point on YP and MLN is Perpendicular to YP. Prove that LM = LN 

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Sol: $\angle YLM=\angle YL N=90^{\circ}$

YL = YL (Common)

$\angle M YL=\angle N YL$ (Bisect by YP )

by ASA $\triangle MY L \cong \triangle N Y L$

By C.P.C.T LM = LN Hence proved 

Question 6

In the fig, PM =PN , PM $\perp A B$ AND $P N \perp A C$ Prove $\triangle A M P \cong \triangle A N P$

Sol: (image to be added)

$\angle A N P=\angle A M P=90^{\circ}$ 

$A P=A P \quad($ Comimon $)$

$M P=N P \quad$ (given)

By R.H.S $\triangle A M P \cong \triangle A N P$ Hence proved 

Question 7

In the figure , AD = BC and AD ||BC. Prove that AB =DC 

Sol: (IMAGE TO BE ADDED)

AD=BC 

$\angle C A D=\angle A C B$ [Alt. angles]

AC = AC 

By ASA $\triangle A C D \cong \triangle A B C$

by C.P.CT. $A B=D C$ Hence proved 

Question 8

In the given figure, triangles ABC and DCB are right angles at A and D respectively and AC = DB . Prove that $\triangle A B C \cong \triangle D C B$

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Sol: $\angle B A C=\angle B D C=90^{\circ}$

$B C=B C=$ Common

$A C=D B$

By RHS $\triangle A B C \cong \triangle D C B$ Hence proved