In an acute - PQR -PQR-45- -Prove that -PQ-2 -PR-2 -QR-2 - 4- area- PQR

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Every Point on the Bisector of an Angle Is Equidistant from the Sides of the Angle.

In an acute △...

Question

In an acute △ PQR ,\angle PQR=45^° .Prove that (PQ)^{2 }=(PR)^{2 }+(QR)^{2 }- 4× area(△ PQR

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△ A B C

△ P Q R

△ A B C \sim △ P Q R

\frac{a r e a (△ A B C)}{a r e a (△ P Q R)} = \frac{A B^{2}}{P Q^{2}} = \frac{B C^{2}}{Q R^{2}} = \frac{A C^{2}}{P R^{2}}

△ P Q R

Q R^{2} = P Q^{2} + P R^{2}

∠ P

Δ A B C

Δ P Q R

A B C \leftrightarrow P Q R

\frac{A (A B C)}{A (P Q R)} = \frac{A B^{2}}{P Q^{2}} = \frac{B C^{2}}{Q R^{2}} = \frac{A C^{2}}{P R^{2}}

△ P Q R, Q M

P R

P R^{2} - P Q^{2} = Q R^{2}

Q M^{2} = P M \times M R

{P R}^{2} + {P Q}^{2} = {Q R}^{2}

90^{o}

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