Question

In the given figure , ∠ADB = ∠BCA and

∠ABD = ∠ BAC. Prove that -

( a ) ΔABD ≅ ΔBAC ( b) AD = BC​

Answer 1

Answer:

Given:- In the given figure,

ADB = BCA

ABD = BAC

(a) To Prove :- ∆ABD ~ ∆BAC

Proof :- In ∆ABD and ∆BAC,

ADB = BCA (given)

ABD = BAC (given)

AB = AB (common)

Therefore, ∆ABD ~ ∆BAC ----- 1

Hence proved.

(b) Since, ∆ABD ~ ∆BAC

By CPCT,

AD = BC

Hence proved.

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Answer 2

Given :-

• ∠ADB = ∠BCA
• ∠ABD = ∠BAC

To Prove OR Show :-

• ΔABD ≅ ΔBAC
• AD = BC

Proof OR Explanation :-

_____________________

$1. \: \bf \bigstar \: In \: \triangle ABD \: and \: \triangle BAC \: \bigstar$

$\rm : \: \longrightarrow \angle ADB = \angle BCA \qquad \bold{[Given]}$

$\rm : \: \longrightarrow \angle ABD = \angle BAC \qquad \bold{[Given]}$

$\rm : \: \longrightarrow AB = BA \qquad \qquad \; \; \; \: \bold{[Common]}$

$\red { \underline { \boxed { \rm \therefore \: \bold {\triangle \: ABD \cong \triangle \: BAC} }}}$

_____________________

$2. \: \bf \bigstar \: To \: Prove \: AD = BC \: \bigstar$

$\rm : \: \longrightarrow \angle ADB = \angle BCA \qquad \bold{[Given]}$

$\rm : \: \longrightarrow \angle ABD = \angle BAC \qquad \bold{[Given]}$

$\rm : \: \longrightarrow AB = BA \qquad \qquad \; \; \; \: \bold{[Common]}$

$\rm Hence, \triangle \: ABD \cong \triangle BAC$

$\red { \underline { \boxed { \rm \therefore \: AD = BC }}} \qquad \bold{[By \: C.P.C.T]}$

_____________________

So, In this way we can prove them.

Hope This Helps You.