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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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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
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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