Question

D is a point on BC such that AB=AC In triangle ABC then show that AB is greater than AB)AD

Answer 1

In any triangle, the side opposite the greater angle is greater than the side opposite the smaller one. This is a consequence of the , and the fact that the sum of the angles of a triangle is π.

Now, in ΔABC, as AB>AC, ∠ACD>∠ABD.

Now, in ΔACD, the exterior angle ∠ADB=∠ACD+∠CAD (sum of interior opposite angles) > ∠ABD+∠CAD>∠ABD.

In ΔADB,∠ADB>∠ABD. Therefore AB>AD.

Answer 2

Step-by-step explanation:

In ∆ABC is the angle opposite to side AB ∠ACB. The angle opposite to side AC is ∠ABC. ... Angles opposite to bigger side is bigger and in simliar way side opposite to bigger angle is larger in the triangle.