+98 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 \(\overline{WY}\), meeting \(\overline{WY}\) at \(O\) and \(P\) as shown. Prove that \(OW=PY\).
+128475 Note that triangle XOY is similar to triangle YPZ .....so....
XO / OY = YP / PZ → XO * PZ = YO * YP
Also, triangle XOW is similar to triangle WPZ ...so....
XO / OW = WP / PZ → XO * PZ = OW * WP
Which implies that
YO * YP = OW * WP
But YO = YP + PO and WP = OW + PO
Therefore....by substitution.....
[ YP + PO ] YP = OW [ OW + PO ] simplify
YP^2 - OW^2 = PO [ OW - YP ] subtract the right side from both sides
[ YP + OW] [ YP - OW ] - PO [ OW -YP] = 0 factor out a negative
[ YP + OW ] [ YP - OW] + PO [ YP - OW] = 0 factor out [ YP - OW]
[ YP - OW] [ YP + OW + PO ] = 0
So, by the zero factor property, either
[ YP + OW + PO ] = 0
But this is impossible since YP, OW and PO are all > 0
Or
[ YP - OW] = 0
Which implies that YP = OW → PY = OW
