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Theorem of Geometric Mean

In Δ ABC,∠ ...

Question

In $Δ A B C, ∠ A C B$ is an abtuse angle. Seg. AD $⊥$ line BC, then show that: $A B^{2} = B C^{2} + A C^{2} + 2 B C . C D$

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Solution

In $△ A D C$
$∠ D = 90^{\circ}$
$A D^{2} + B D^{2} = A C^{2}$
In $△ A D B$
$∠ D = 90^{\circ}$
$A D^{2} + B D^{2} = A B^{2}$
As we know that $B D = B C + C D$
So, $A B^{2} = A d^{2} + (B C + C D)^{2}$
$⟹ A B^{2} = A C^{2} + B C^{2} + 2 B C + C D$
Hence proved.

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Similar questions

△ A B C, ∠ B

A D ⊥ B C

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

Δ A B C, ∠ A = 90^{\circ}

A D ⊥ B C .

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

Δ A B C

A D ⊥ C B

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

△ A B C

⊥

A C^{2} = A D^{2} + B C \times D M + \frac{B C^{2}}{4}

A B^{2} = A D^{2} - B C \times D M + \frac{B C^{2}}{4}

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