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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 \( \angle PEQ = 60^\circ \).

Note by Vishwathiga Jayasankar
9 months, 1 week 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.

\(\angle OPX=\angle OXP, \angle OQY=\angle OYQ\)

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

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

\(\angle E=\angle OXY\) Maria Kozlowska · 9 months ago

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Akshat Sharda · 9 months, 1 week ago

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Are PQ and XY parallel? Deeparaj Bhat · 9 months, 1 week ago

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@Deeparaj Bhat 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 · 9 months, 1 week ago

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@Brian Charlesworth So it will be 60° even if PQ and XY are not parallel ? Akshat Sharda · 9 months, 1 week ago

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@Akshat Sharda Yes, I haven't determined a proof yet, but that result does seem to hold in general. Brian Charlesworth · 9 months ago

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