Question

If D is a point on side BC of a A ABC such that AD = BD
= CD, then
(A) AB2 = BC2 + CA2
(B) BC2 = AB2 + CA2
(C) CA2 = AB2 + BC2
(D) AB.AC = AD2
which is correct and why​

Answer 1

Answer:

Step-by-step explanation:

Given: AB = BC = CA = x

∠ADC = ∠ADB = 90°

To prove: AB2 + BC2 + CA2 = 4AD2

Proof:

We know that in an equilateral triangle perpendicular from any vertex on opposite side bisects it.

So, BD = DC = 1/2BC ……………… (1)

By applying Pythagoras theorem in ΔABD, we get,

AB2 = AD2 + BD2 [∵ H2 = P2 + B2]

By substituting BD from eqn. (1) we get,

AB2 = AD2 + (1/2BC)2

⇒ AB2 = AD2 + 1/4BC2

⇒ 4AB2 = 4AD2 + BC2

⇒ 4AB2 - BC2 = 4AD2

⇒ 4AB2 - AB2 = 4AD2 [Given: AB = BC]

⇒ 3AB2 = 4AD2

⇒ AB2 + BC2 + CA2 = 4AD2 [∵ AB = BC = CA]

Hence proved.