In triangle ABC, if $A P ⊥ B C$ and $A C^{2} = B C^{2} - A B^{2}$ , then prove that $P A^{2} = P B \times C P$ .

Open in App

Solution

$A C^{2} = B C^{2} - A B^{2}$

[Given]

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

$∴ ∠ B A C = 90^{\circ}$

[By converse of Pythagoras' theorem]

$Δ A P B \sim Δ C P A$

[If a perpendicular is drawn from - the vertex of the right angle of a triangle to the hypotenuse then triangles on both sides of the perpendicular are similar to the whole triangle and to each other ]

$\Rightarrow \frac{A P}{C P} = \frac{P B}{P A}$

[In similar triangle, correpsonding sides are proportional]

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

Suggest Corrections

Similar questions

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

A B C

∠ A B C < 90^{o}

A D ⊥ B C .

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

A B C

∠ A B C < 90^{\circ}

A D ⊥ B C .

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

△ A B C, ∠ B

A D ⊥ B C

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

In fig., $A B C$ is a triangle in which $∠ A B C > 90^{o}$ and $A D ⊥ C B$ produced. Prove that $A C^{2} = A B^{2} + B C^{2} + 2 B C . B D$ .

Join BYJU'S Learning Program