21. If two isosceles triangles have a common base, the line joining th

1.	21. If two isosceles triangles have a common base, the line joining their vertices bisects themat right angles.
Answer» Given:ΔABC and ΔDBC are isosceles triangle with common base BC. AB = AC and BD = DC. To prove:AD bisects BC at 90°. Proof: In ΔABD and ΔACD, AB = AC (Given) BD = CD (Given) AD = AD (Common) ∴ ΔABDΔACD (SSS Congruence criterion) ⇒ ∠1 = ∠2 ...(1) (CPCT) In ΔABE and ΔACE, AB = AC (Given) ∠1 = ∠2 [Using (1)] AE = AE (Common) ∴ ΔABEΔACE (SAS congruency criterion) ⇒ BE = CD ...(2) (CPCT) and ∠3 = ∠4 (CPCT) Now, ∠3 + ∠4 = 180° (Linear pair ) ∴ 2∠3 = 180° (∠3 = ∠4) ⇒ ∠3 = 90° ...(3) Hence, AD bisects BC at 90° [Using (2) and (3)]

21. If two isosceles triangles have a common base, the line joining their vertices bisects themat right angles.

Answer»

Given:ΔABC and ΔDBC are isosceles triangle with common base BC. AB = AC and BD = DC.

To prove:AD bisects BC at 90°.

Proof:

In ΔABD and ΔACD,

AB = AC (Given)

BD = CD (Given)

AD = AD (Common)

∴ ΔABDΔACD (SSS Congruence criterion)

⇒ ∠1 = ∠2 ...(1) (CPCT)

In ΔABE and ΔACE,

AB = AC (Given)

∠1 = ∠2 [Using (1)]

AE = AE (Common)

∴ ΔABEΔACE (SAS congruency criterion)

⇒ BE = CD ...(2) (CPCT)

and ∠3 = ∠4 (CPCT)

Now, ∠3 + ∠4 = 180° (Linear pair )

∴ 2∠3 = 180° (∠3 = ∠4)

⇒ ∠3 = 90° ...(3)

Hence, AD bisects BC at 90° [Using (2) and (3)]