In triangle $A B C, A B = A C$ and $B D$ is perpendicular to $A C$ . Prove that:
$B D^{2} - C D^{2} = 2 (C D \times A D)$

Open in App

Solution

ABC is a isosceles triangle in which $A B = A C$
$B D$ Perpendicular to $A C$
In $△ A B D$
${A B}^{2} = {A D}^{2} + {B D}^{2}$ ...(Pythagoras theoram)
$∴ {A C}^{2} = {A D}^{2} + {B D}^{2}$ ....(as AB=AC) ...(1)
From diagram, we see that
$\Rightarrow A C = C D + A D$
$\Rightarrow {A C}^{2} = {(C D + A D)}^{2}$
From (1) we get,
$\Rightarrow {(C D + A D)}^{2} = {A D}^{2} + {B D}^{2}$
${C D}^{2} + {A D}^{2} + 2 (C D \times A D) = {A D}^{2} + {B D}^{2}$
${C D}^{2} + 2 (C D \times A D) = {B D}^{2}$ ...(AD square gets cancelled from both sides)
${B D}^{2} - {C D}^{2} = 2 (C D \times A D)$
Hence proved.

Suggest Corrections

Similar questions

Δ

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

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

Join BYJU'S Learning Program