In $△$ ABC, given that AB=AC and $B D ⊥ A C$ . Prove that ${B C}^{2} = 2 A C \cdot C D$

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Solution

From $Δ$ ABD we have

$A B^{2} = B D^{2} + A D^{2}$ ........(1). [By Pythagoras Theorem]

And from $Δ$ BCD we have

$B C^{2} = B D^{2} + C D^{2}$ .........(2). [By Pythagoras Theorem]

Subtracting (2) from (1) we get

$\Rightarrow A B^{2} - B C^{2} = A D^{2} - C D^{2}$

$\Rightarrow A B^{2} = B C^{2} + (A D + C D) (A D - C D)$

$\Rightarrow A B^{2} = B C^{2} + A C (A D - C D)$

Given $A B = A C$ ;

$\Rightarrow A C^{2} = B C^{2} + A C . A D - A C . C D$

$\Rightarrow A C (A C - A D) = B C^{2} - A C . C D$

$\Rightarrow A C . C D = B C^{2} - A C . C D$

$\Rightarrow B C^{2} = 2 A C . C D$ $[h e n c e p r o v e d]$

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In the given figure, $△ A B C$ is an isosceles triangle in which $A B = A C .$ If $B D ⊥ A C$ and $C E ⊥ A B$ , prove that $B D = C E .$

Δ

∠

⊥

⊥

B C^{2} = (A C \times C P + A B \times B Q)

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