Question

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In ∆ABC points D and E are on BC,
Such that BD = EC and AD = AE prove that AB = AC.

Answer 1

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Answer 2

$\huge\bigstar\underline\mathfrak\pink{Solution}$

Given : In ∆ABC points D and E are on BC,

Such that BD = EC and AD = AE.

To prove : AB = AC

Proof : Angle 1 + Angle 3 = 180° ( Linear pair )...(i)

Similarly, Angle 4 + Angle 2 = 180° ( Linear pair )....(ii)

But, Angle 3 = Angle 4 ( Because AD = AE and we know that angles opposite to equal sides are equal )...(iii)

From (i), (ii), (iii)

Angle 1 + Angle 3 = Angle 4 + Angle 2 ( Angles equal to the same angle are equal to one another )

Angle 3 and Angle 4 are cancelled out as both are equal [ from (iii) ]

Therefore, Angle 1 = Angle 2

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Now, In ∆ADB and ∆AEC

AD = AE ( Given )

Angle 1 = Angle 2 ( proved above )

BD = EC ( Given )

Hence, ∆ADB is congruent to ∆AEC

So, AB = AC ( corresponding parts of congruent triangles are equal )

________________________

Hence proved!