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# More Right Triangles!

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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 $$OW=PY$$.

benjamingu22  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

CPhill  Sep 15, 2017
edited by CPhill  Sep 15, 2017

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