Geometry Proofing Problem

Let ABCD be a rectangle and let P be a point on it circumcircle, different from any vertex. Let X, Y, Z and W be the projections of P onto the lines AB, BC, CD, and DA, respectively. Prove that one of the points X, Y, Z and W is the orthocenter of the triangle formed by the other three.

#Geometry #Proofs #MathProblem #Math

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

Hint: Lines $AB$ and $CD$ are perpendicular if and only if $AC^2 + BD^2 = AD^2 + BC^2$ .

This is one of my favorite ways of showing that 2 lines are perpendicular.

Using the above hint, draw a diagram, and the result is almost immediate.

Calvin Lin Staff - 7 years, 10 months ago

Thanks you (y)

Dani Natanael - 7 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$