Question

31. triangleABC and triangle
ADC are two isosceles triangles on the same base BC and vertices A and D
are on the same side of BC. I AD is extended to intersect BC at P, show that
(і) triangleАBD≈triangleACD

Answer 1

Proved that ΔABD ≅ ΔACD.

To prove : ΔABD ≅ ΔACD.

Given :

ΔABC and ΔADC are two isosceles triangle.  

It lies on the same base BC and vertices A and D.

From the figure,

Note : Figure is attached below.

In ΔABC,

AB = AC

Here, ΔABC is isosceles.

In ΔBDC,

BD = DC

Here, ΔBDC is isosceles.

AD = AD is common.

By applying, SSS congruent rule

ΔABC ≅ ΔACD.

Hence proved that ΔABC ≅ ΔACD.    

To learn more...

Prove that the sum of all angles of a triangle is 180 . Also, find the angles of a triangle if they are in ratio 5:6:7.

brainly.in/question/1212164