Question

prove that the altitude from vertex of an isosceles triangle bisect the base​

Answer 1

Answer:

Step-by-step explanation:

By RHS Rule of Congruency,

1. The altitude forms the side which is same

2. The hypotenuse of both the triangles formed is same as it is siosceles

3. A 90 degree angle is formed at the point where the altitude strikes the       base which is common

Therefore, the base gets divided into two equal parts, bisecting it.

Answer 2

Let, ABC is a ∆.

To prove that :- BD = CD

Construction:- Draw a line segment perpendicular to BC.

Proof:- In ∆ADB and ∆ADC.

<B = <C ( base angles are equal)

<D = <D = 90°

AB = AC (opposite of of isosceles ∆)

Hence, $∆ADB\cong ∆ADC$ (AAS)

Now,

BD = CD (C.P.C.T)