Question

In the given figure ac=ae,ab=ad and angle bad =angle eac. show that bc=de​

Answer 1

Given:-

• AC = AE & AB = AD

• $\sf{\angle{BAD} = \angle{EAC}}$

To Show:-

• BC = DE

Congurency Used:-

• S - A - S → Side angle Side.

Now,

→ $\sf{\angle{BAD} = \angle{EAC}}$

→ $\sf{ \angle{ACD}\:is\:Common}$

• Adding it on both the Sides

→ $\sf{\angle{BAD} +\angle{ACD} = \angle{EAC} +\angle{ACD} }$

→ $\sf{ \angle{BAC} = \angle{EAD}}$...... 1

Therefore,

In ∆BAC and ∆EAD

→ $\sf{ AC = AE }$ ( Given )

→ $\sf{ AB = AD }$ ( Given )

→ $\sf{ \angle{BAC} = \angle{ EAD}}$ ( From 1. )

Hence, ∆BAC ≈ ∆EAD by S - A - S Congurency Criteria.

Therefore,

→ BC = DE by C.P.C.T

Answer 2

Given:-

◍ AC = AE

◍ AB = AD

◍ ∠BAD = ∠EAC

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To prove:-

◍ BC = DE

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Solution:-

As, ∠BAD=∠EAC

➝ ∠BAD+∠ACD = ∠EAC+∠ACD (∠ACD is common)

➝ ∠BAC=∠EAD -(i)

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In ∆BAC and ∆DAE

AC=AE [Given]

AB=AD [Given]

∠BAC=∠DAE [from (i)]

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By S-A-S congruency criteria

∆BAC ≈ ∆DAE

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By C.P.C.T

➝ BC = DE

${\fcolorbox{white}{lightgreen}{{\footnotesize{{ \color{white}{\checkmark} \: }}}{\huge{Hence, \: proved}}}}$