Question

show that in an isocseles triangle, the angle opposite to equal sides are equal. ​

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 )

Answer 2

Step-by-step explanation:

let triangle ABC be an isosceles triangle with height AD

in triangle ACD & triangle ABD

AD =AD(common)

AB=AC (isosceles triangle)

ANGLE ADC = ANGLE ADB (90° AD perpendicular to BC)

so triangle ABD is congruent to triangle ACD

ANGLE ACD = ANGLE ABD (CPCT)

hence proved