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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 \(\overline{WY}\), meeting \(\overline{WY}\) at \(O\) and \(P\) as shown. Prove that \(OW=PY\).

 

 Sep 14, 2017
edited by benjamingu22  Sep 14, 2017
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 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

 

 

cool cool cool

 Sep 15, 2017
edited by CPhill  Sep 15, 2017

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