Question

Prove that angles opposite to equal sides of an isosceles triangle are equal.

please give the proof ​

Answer 1

Answer:

Let △ABC be an isosceles triangle such that AB =AC Then we have to prove that ∠B=∠C Draw the bisector AD of ∠A meeting BC in D

Now in triangles ABD and ACD We have AB=AC (Given)

∠BAD=∠CAD (because AD is bisector of ∠A

AD=AD (Common side)

Therefore by SAS congruence condition we have

△ABC≅△ACD

⇒∠B=∠C

(Corresponding parts of congruent triangles are equal ).

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Answer 2

$\huge \sf { \fcolorbox{green}{g}{❃ \: Explanatation \: ❃}}$

$\Large{\underline{ \underline{\sf{Given:}}}}$

Isosceles triangle ∆ ABC

i. e, AB = AC

$\Large{\underline{ \underline{\sf{To \: prove:}}}}$

∠ B = ∠ C

$\Large{\underline{ \underline{\sf{Construction:}}}}$

Draw a bisector of ∠A intersecting BC at D.

$\Large{\underline{ \underline{\sf{Proof:}}}}$

In ∆ BAD and ∆CAD

AB = AC ( given )

∠BAD = ∠CAD ( by construction )

AD = AD ( common )

∆ BAD ≅ ∆ CAD

( By SAS Congruence rule )

Thus,

∠ABD = ∠ACD ( By CPCT )

∠B = ∠C

Hence, Angle opposite to equal sides are equal.

Hence, Proved