Question

angles opposite to equal sides of an isosceles triangle are equal prove that

Answer 1

in triangle ABC

AB=AC

DRAW A LINE 'AD' SUCH THAT IT MEETS A IN WHICH AD⊥ BC

NOW PROVE ABD≅ACD

IN ΔABD AND ΔACD

∠ADB=∠ADC=90°

AB=AC (given)

AD=AD (common)

hence ΔADB≅ΔADC by RHS Cong. rule

∴∠ABD=∠ACD (CPCT)

∴ Angles opposite to equal sides are equal

HENCE PROOOVED!!!!!!

Answer 2

Consider ΔABC an isosceles triangle

⇒ AB=AC

Construction: AD is bisector on BC

So, ∠BAD=∠CAD

In △ABD and △ACD

⇒ AB=AC

⇒ ∠BAD=∠CAD

⇒ AD=AD

So, △ABD≅△ACD ..........(SAS)

hence, ∠ABC=∠ACB