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Pythagoras Theorem

The perpendic...

Question

The perpendicular from A on side BC of a $Δ A B C$ intersects BC at D such that DB = 3CD .Prove that $2 A B^{2} = 2 A C^{2} + B C^{2}$ .

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Solution

Given that in $Δ A B C$ , we have
$A D ⊥ B C a n d B D = 3 C D$
In right angle triangles ADB and ADC, we have
$A B^{2} = A D^{2} + B D^{2} . . . (i)$
$A C^{2} = A D^{2} + D C^{2} . . . (i i)$ [By Pythagoras theorem]
Subtracting equation (ii) from equation (i), we get
$A B^{2} - A C^{2} = B D^{2} - D C^{2}$
$= 9 C D^{2} - C D^{2} [∴ B D = 3 C D]$
$= 8 C D^{2}$
$= 8 {(\frac{B C}{4})}^{2} [S i n c e, B C = D B + C D = 3 C D + C D = 4 C D]$
Therefore, $A B^{2} - A C^{2} = \frac{B C^{2}}{2}$
$\Rightarrow 2 (A B^{2} - A C^{2}) = B C^{2}$
$\Rightarrow 2 A B^{2} - 2 A C^{2} = B C^{2}$
$∴ 2 A B^{2} = 2 A C^{2} + B C^{2}$

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