Question

ABC is a triangle. AD perpendicular BC and BD = 3 CD. Prove that 2(AB)2 = 2(AC)2 + BC2​

Answer 1

Proved below.

Step-by-step explanation:

Given:

DB = 3CD                          [1]

As shown in the figure, BC = CD + DB  

BC = CD + 3CD                 [from 1]

⇒ BC = 4CD                     [2]

Here it is given that  AD ⊥ BC, so

In Δ ACD

By Pythagoras theorem  

$AC^2 = AD^2 + CD^2$           [3]

In ΔABD

$AB^2 = AD^2 + DB^2$

⇒ $AB^2 = AD^2 + (3CD)^2$

⇒ $AB^2 = AD^2 + 9CD^2$     [4]

Subtracting [3] from [4]

⇒$AB^2 - AC^2 = AD^2 + 9CD^2 - (AD^2 + CD^2 )$

⇒ $AB^2 - AC^2 = AD^2 + 9CD^2 - AD^2 - CD^2$  

⇒ $AB^2 - AC^2 = 8CD^2$  

⇒ $AB^2 = 8CD^2 + AC^2$

⇒ $AB^2 = AC^2 + 8CD^2$

Multipying both the side by 2, we get

⇒ $2AB^2 = 2AC^2 + 16CD^2$

⇒$2AB^2 = 2AC^2 + (4CD)^2$

⇒ $2AB^2 = 2AC^2 + BC^2$         [From 2]

Hence Proved.