In $Δ$ ABC, AD $⊥$ BC and point D lies on BC such that $2$ DB $= 3$ CD. Prove that $5 A B^{2} = 5 A C^{2} + B C^{2}$ .

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Solution

Given : $A D ⊥ B C$

$2 D B = 3 C D$

$\frac{D B}{C D} = \frac{3}{2}$

$\Rightarrow D B = 3 x, C D = 2 x$ and $B C = 5 x$

$ln Δ A D B, ∠ B D A = 90^{\circ}$

$A B^{2} = A D^{2} + D B^{2}$

$A B^{2} = A D^{2} + (3 x)^{2}$

$A B^{2} = A D^{2} + 9 x^{2}$

$5 A B^{2} = 5 A D^{2} + 45 x^{2}$

$5 A D^{2} = 5 A B^{2} - 45 x^{2}$ (i)

and $A C^{2} = A D^{2} + C D^{2}$

$A C^{2} = A D^{2} + 4 x^{2}$

$5 A C^{2} = 5 A D^{2} + 20 x^{2}$

$5 A D^{2} = 5 A C^{2} - 20 x^{2}$ (ii)

comparing (i) and (ii)

$5 A B^{2} - 45 x^{2} = 5 A C^{2} - 20 x^{2}$

$5 A B^{2} = 5 A C^{2} + 25 x^{2}$

$= 5 A C^{2} + (5 x)^{2}$

$5 A B^{2} = 5 A C^{2} + B C^{2}$

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Similar questions

Δ

⊥

= 3

2 A B^{2} = 2 A C^{2} + B C^{2}

The perpendicular from A on side BC of a ΔABC intersect BC at D such that DB = 3 CD. Prove that 2 AB² = 2 AC² + BC²

△ A B C

D B = 3 \times C D

2 {A B}^{2} = 2 {A C}^{2} + {B C}^{2}

DC = \frac{1}{4} BC

Δ A B C

2 A B^{2} = 2 A C^{2} + B C^{2}

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