prove that angle opposite to equal sides of an isosceles triangle are equal
Answers
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