Can you please help me solve this?

$PQ$ is a diameter of circle and $XY $ is chord equal to the radius of the circle. $PX$ and $QY$ when extended intersect at $E$. Prove that $ \angle PEQ = 60^\circ $.

#Geometry

Easy Math Editor

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Comments

Let $O$ denote circle center. It can be shown in general that $\angle PEQ = \angle OXY=\angle OYX$ regardless of $XY$ length and if it is parallel to $PQ$ or not.

$\angle OPX=\angle OXP, \angle OQY=\angle OYQ$

$\angle E = 180-\angle OPX - \angle OQY = 180-\angle OXP-\angle OYQ=180 - \angle YXE - \angle XYE$

$(180-\angle OXP - \angle YXE) +(180-\angle OYQ-\angle XYE)=2 \angle OXY = 2 \angle E \Rightarrow$

$\angle E=\angle OXY$

Maria Kozlowska - 5 years, 3 months ago

Are PQ and XY parallel?

A Former Brilliant Member - 5 years, 3 months ago

I at first thought that might be a necessary condition, but after looking at several orientations for $XY$ it appears that $\angle PAQ = 60^{\circ}$ in general, which would be an interesting result.

Brian Charlesworth - 5 years, 3 months ago

So it will be 60° even if PQ and XY are not parallel ?

Akshat Sharda - 5 years, 3 months ago

@Akshat Sharda – Yes, I haven't determined a proof yet, but that result does seem to hold in general.

Brian Charlesworth - 5 years, 3 months ago

Akshat Sharda - 5 years, 3 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}`	$\boxed{123}$