Required to prove :-

Angles opposite to equal sides of an isosceles triangle are equal.

Construction :-

Join A to D such that BD = CD which makes AD as the altitude .

Proof :-

Diagram :-

Consider ∆ ABC ;

∆ ABC is an isosceles triangle .

In which , AC = AB

we need to prove that ;

In order to prove this . First we need to prove that the two TRIANGLES i.e. ∆ ADC & ∆ ADB are CONGRUENT with each other .

So,

Consider ∆ ADC & ∆ ADB

In ∆ ADC & ∆ ADB

AD = AD

[ Reason :- Common Side ]

CD = DB

[ Reason :- By construction ]

AC = AB

[ Reason :- Given information ]

So,

By SSS congruency criteria

we can say that ;

∆ ADC ∆ADB

This implies ;

[ Reason :- CORRESPONDING parts of congruent triangles ]

Hence,

It is PROVED that the angles opposite to equal sides of an isosceles triangle are equal .

Additional Information :-

The CONVERSE of the above theorem is ;

If the opposite angles of an isosceles triangle are equal then the opposite sides are also equal .

While solving these proof. The congruency rules are ver useful .

Some of them are ;

• SSS ( Side , Side , Side )

• AAA ( Angle , Angle , Angle )

• SAS ( Side , Angle , Side )

• ASA ( Angle , Side , Angle )

• RHS ( Right angle , Hypotenuse , Side )

Here,

AAA rule is not mostly accepted rule .

Some of the important theorem which are useful while solving these proofs are ;

• Thales theorem

• Mid point theorem