In any $△ A B C, ∠ B$ is an acute angle and $A D ⊥ B C$ , then prove that:
$A C^{2} = A B^{2} + B C^{2} - 2 B C . B D$ .

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Solution

In $△ A B C, ∠ A B D = 90 °$
By pathagores theorem
${A B}^{2} = {A D}^{2} + B D^{2}$
In $△ A D C, ∠ A D C = 90 °$
${A C}^{2} = {A D}^{2} + D C^{2}$
${A C}^{2} = {A D}^{2} + (B C - B D)^{2}$
${A C}^{2} = {A D}^{2} + B C^{2} + B D^{2} - B C \times B D$
${A C}^{2} = {(A D}^{2} + B D^{2}) + B C^{2} - B C \times B D$
${A C}^{2} = {A B}^{2} + B C^{2} - B C \times B D$
hence proved

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In the given figure, ABC is a triangle in which ∠ABC < 90° and AD ⊥ BC. Prove that AC² = AB² + BC² − 2BC.BD.

In the given figure, ABC is a triangle in which ∠ABC> 90° and AD ⊥ CB produced. Prove that AC² = AB² + BC² + 2BC.BD.

△ A B C, ∠ C

A E ⊥ B C

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

△ A B C

A D

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

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∠ A B C < 90^{o}

A D ⊥ B C .

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

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