Question

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

Answer 1

Answer:

ABC is a given triangle with, AB=AC.

To prove: Angle opposite to AB= Angle

opposite to AC (i.e) ∠C=∠B

Construction: Draw AD perpendicular to BC

∴∠ADB=∠ADC=90

o

Proof:

Consider △ABD and △ACD

AD is common

AB=AC

∴∠ADB=∠ADC=90

o

Hence ∠ABD=∠ACD

∠ABC=∠ACB

∠B=∠C. Hence the proof

Answer 2

Step-by-step explanation:

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 )