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**lovelylola****Member**- Registered: 2014-07-09
- Posts: 2

Quadrilateral WXYZ has right angles at angle W and angle Y and an acute angle at angle X. Altitudes are dropped from X and Z to diagonal WY, meeting WY at O and P. Prove that OW = PY.

My name is Lola, and indeed I am quite lovely.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,426

Hi Lola,

Welcome to the forum.

Just got back from a holiday, so I've only just seen this.

Note: WXYZ is a cyclic quadrilateral. ie. There is a circle (let's say centre C) that goes through all four points.

If you extend ZP until it cuts the circle again at Q; and extend XO until it cuts the circle again at R,

then

QZ is parallel to XR, and CQ = CZ = CR = CX

so

QP = XO and PZ = OR.

There is a property of any two chords in a circle that

WP.PY = QP.PZ

and

OY.OW = XO.OR

Thus

WP.PY = OY.OW => WP(PO+OY) = OY(WP+PO)

=> WP.PO + WP.OY = OY.WP + OY.PO

Simplify to WP = OY and the required result follows.

If you've not met any of those theorems for a circle, post back.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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