Question

Prove that the straight line joining the vertex of an isosceles triangle to any point in the base is smaller than either of the equal sides of the triangle

Answer 1

Answer:

True

Step-by-step explanation:

asumming the length of base be 'b', height be 'h' and equal sides be 'l'

since l*2 =h*2 + (b/2)*2

let's take any 'x' distance from centre of base then the length of straight line will be (assuming 'y') y*2= h*2 + x*2

here x< b/2 hence y<l

Answer 2

Answer:

We know that the exterior angle of a triangle is always greater than each of the interior opposite angles.

Step-by-step explanation:

∴, In ΔABD,

∠ADC >∠B ------- (1)

In ΔABC,

AB=AC

∴, ∠B =∠C (Converse of isosceles triangles theorem) ------- (2)

In ΔADC,

From (1) and (2),

∠ADC >∠C

∴, AC >AD (side opposite to greater angle is greater) ------(3)

In ΔABC,

∵, AB=AC

AB >AD

Hope this helps:)