In $Δ A B C, ∠ A C B$ is obtuse. $A D ⊥ B C$ produced and $B E ⊥ A C$ produced. Prove that $A B^{2} = B C . B D + A C . A E$

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Solution

Given in $△ A B C, ∠ A$ is obtuse and $A D ⊥ C B$ and $B E ⊥ C A$

Consider $△ A D C a n d △ C E B$

$∠ D = ∠ E = 90^{\circ}$

$∠ D C A = ∠ E C B$ [Vertically opposite angles]

$\Rightarrow △ A D C \sim △ C E B$ [AAA similarity criterion]

Consider, right angled triangle ABE

$\Rightarrow A B^{2} = B E^{2} + A E^{2}$ [By Pythagoras theorem]

$\Rightarrow A B^{2} = B E^{2} + (C A + C E)^{2}$ [Since AE = AC + CE]

$\Rightarrow A B^{2} = [B E^{2} + C E^{2}] + A C^{2} + 2 A C \times C E$ (2)

In right $△ B C E, B E^{2} + C E^{2} = B C^{2}$ [By Pythagoras theorem]

Hence equation (2) becomes,

$\Rightarrow A B^{2} = B C^{2} + A C^{2} + C A \times C E + C A \times C E$

$\Rightarrow A B^{2} = B C^{2} + A C^{2} + C A \times C E + C D \times C B$ [From (1)]

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

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

$\Rightarrow A B^{2} = B C \times B D + A C \times A E$ [From the figure]

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