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)
AD = AD (Common)
∴ 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!