In triangle ABC, AC=AB and BD is perpendicular to AC prove that $B D^{2} + C D^{2} = 2 A C \times C D$

Open in App

Solution

Given: BD is perpendicular to AC and AB = AC
To prove: $B C^{2} = 2 A C . C D$
Proof: In $Δ B C D$ by pythagoras theorem, we have
$\Rightarrow B C^{2} = B D^{2} + C D^{2} \dots \dots (1)$
Again, in $Δ A B D$ by pythagoras theorem
$A B^{2} = B D^{2} + A D^{2}$
$\Rightarrow B D^{2} = A B^{2} - A D^{2}$
On putting value of $B D^{2}$ in (1), we get
$B C^{2} = A B^{2} - A D^{2} + C D^{2}$
$\Rightarrow B C^{2} = A B^{2} - (A C - C D)^{2} + C D^{2}$
$\Rightarrow B C^{2} = A B^{2} - A C^{2} - C D^{2} + 2 A C . C D + C D^{2}$
$\Rightarrow B C^{2} = 2 A C . C D [G i v e n, A B = A C \Rightarrow A B^{2} = A C^{2}]$

Suggest Corrections

Similar questions

A B C

A B = A C

B D

A C

{B D}^{2} - {C D}^{2} = 2 C D \times A D

A B C

A B = A C,

B D

B

A C .

B D^{2} - C D^{2} = 2 C D . A D

B D^{2} - C D^{2} = 2 C D . A D

Δ

B D^{2} - C D^{2} = 2 C D \times A D

Join BYJU'S Learning Program