Question

in the given figure angle abd = ebc bd = bc angle acb = angle edb prove ab = be​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Given,

In Δ$ABC$ and Δ$EBD$,

∠$ABD$ = ∠$EBC$

$BD$ = $BC$

∠$ACB$ = ∠$EDB$

Add ∠$ABE$ to both sides of ∠$ABD$ = ∠$EBC$, after adding we get,

∠$ABE$ + ∠$ABD$ = ∠$EBC$ + ∠$ABE$

∠$ABC$ = ∠$EBD$

$DB$ = $BC$

∠$ACB$ = ∠$EDB$

So by Angle-Side-Angle (ASA) Congruency, we prove that

Δ$ABC$ ≅ Δ$EBD$

∴ $AB$ = $BE$ ( By Corresponding parts of a congruent triangle)

Hence Proved.

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