Question

please tell me this answer​

Answer 1

TO PROVE :-

• The angles opposite to the equal sides of an isosceles triangle are equal. (By drawing altitude through the vertex to the base).

TO FIND :-

• Congruence criterion used to prove the required result.

SOLUTION :-

Refer the attachment.

Let AD be the altitude of the isosceles triangle ABC having base angles ABC and ACB equal.

Here , AB and AC are opposite sides to the equal angles. So, we have to prove that AB = AC.

Altitude bisects the base and the corresponding angle.

Therefore ,

• BD = DC --------(i)
• ∠BAD =∠DAC --------(ii)

In ∆ABD and ∆ACD ,

• ∠ ABD =∠ ACD (given)
• BD = DC ( by equation i )
• ∠BAD =∠DAC ( by equation ii )

Hence , ∆ABD ≌ ∆ACD (By AAS test)

So , all corresponding sides and angles of the triangles are equal.

Therefore , AB = AC (proved)

Hence , angles opposite to the equal sides of an isosceles triangle are equal . We proved this by AAS test.

NOTE :-

• AAS test : Two corresponding angles and one corresponding side of the triangles should be equal.

OTHER CONGRUENCY TESTS :-

• SSS test : All corresponding sides of the trianles are equal.
• RHS test : Two corresponding sides of right angled triangles are equal.
• ASA test :- Two corresponding angles and one side of the triangles is equal.
• SAS test : Two corresponding sides and one corresponding angle of the triangles is equal.