# 1.if two isosceles triangle have a common base,prove that the line segment joining their vertices bisects the common base at right angles.2.show that the sum of three medians of a triangle is less than its perimeter.

Dear Student!

(1) 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)

∴ ΔABD ΔACD     (SSS Congruence criterion)

⇒ ∠1 = ∠2        ...(1)     (CPCT)

In ΔABE and ΔACE,

AB = AC                                 (Given)

∠1 = ∠2                                  [Using (1)]

AE = AE                                  (Common)

∴ ΔABE ΔACE                  (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)]

2. Cheers!

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