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$$PQ$$ 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 $$\angle PEQ = 60^\circ$$.

Note by Vishwathiga Jayasankar
1 year, 8 months ago

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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.

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

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

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

$$\angle E=\angle OXY$$

- 1 year, 8 months ago

- 1 year, 8 months ago

Are PQ and XY parallel?

- 1 year, 8 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.

- 1 year, 8 months ago

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

- 1 year, 8 months ago

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

- 1 year, 8 months ago