Question
In ∆ ABC, seg AD $\perp$ seg BC DB = 3CD. Prove that : $2 A B^2=2 A C^2+B C^2$
✓
Answer
Given,
$D B=3 C D$
Also,
$B C=C D+D B=C D+3 C D$
$\Rightarrow B C=4 C D[1]$
As, $A D \perp B C$, By Pythagoras theorem i.e.
$(\text { Hypotenuse })^2=(\text { base })^2+(\text { Perpendicular })^2$
In $\triangle A C D$
$A C^2=A D^2+C D^2[2]$
In $\triangle A B D$
$A B^2=A D^2+D B^2$
$\Rightarrow A B^2=A D^2+(3 C D)^2$
$\Rightarrow A B^2=A D^2+9 C D^2[3]$
Subtracting [2] from [3]
$\Rightarrow A B^2-A C^2=9 C D^2-C D^2$
$\Rightarrow A B^2=A C^2+8 C D^2$
$\Rightarrow 2 A B^2=2 A C^2+16 C D^2$
$\Rightarrow 2 A B^2=2 A C^2+(4 C D)^2$
$\Rightarrow 2 A B^2=2 A C^2+B C^2[\text { From 1] }$
Hence Proved.
$D B=3 C D$
Also,
$B C=C D+D B=C D+3 C D$
$\Rightarrow B C=4 C D[1]$
As, $A D \perp B C$, By Pythagoras theorem i.e.
$(\text { Hypotenuse })^2=(\text { base })^2+(\text { Perpendicular })^2$
In $\triangle A C D$
$A C^2=A D^2+C D^2[2]$
In $\triangle A B D$
$A B^2=A D^2+D B^2$
$\Rightarrow A B^2=A D^2+(3 C D)^2$
$\Rightarrow A B^2=A D^2+9 C D^2[3]$
Subtracting [2] from [3]
$\Rightarrow A B^2-A C^2=9 C D^2-C D^2$
$\Rightarrow A B^2=A C^2+8 C D^2$
$\Rightarrow 2 A B^2=2 A C^2+16 C D^2$
$\Rightarrow 2 A B^2=2 A C^2+(4 C D)^2$
$\Rightarrow 2 A B^2=2 A C^2+B C^2[\text { From 1] }$
Hence Proved.
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