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 July 9th, 2014, 09:13 PM #1 Newbie   Joined: Jul 2014 From: Mars Posts: 1 Thanks: 0 Right Triangles Proof 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.
 July 10th, 2014, 07:51 PM #2 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,302 Thanks: 935 Changing your lettering to (hate using O and Z!): Quadrilateral ABCD has right angles at A and C and an acute angle at angle B. Altitudes are dropped from B and D to diagonal AC, meeting AC at E and F. Prove that AE = CF. That's same as proving that AF = CE (since EF is common to both AE and CF). a = AB, b = BC, c = CD, d = DA. e = CE, f = AF g = BD, h = AC (diagonals) m = BE, n = DF (heights) Prove that e = f. triangleABD: g^2 = a^2 + d^2 triangleBCD: g^2 = b^2 + c^2 a^2 + d^2 = b^2 + c^2 a^2 + d^2 - b^2 - c^2 = 0 [1] triangleBCE: m^2 = b^2 - e^2 triangleABE: m^2 = a^2 - (h - e)^2 b^2 - e^2 = a^2 - (h - e)^2 Simplifies to: h^2 = a^2 - b^2 + 2eh [2] triangleADF: n^2 = d^2 - f^2 triangleCDF: n^2 = c^2 - (h - f)^2 d^2 - f^2 = c^2 - (h - f)^2 Simplifies to: h^2 = c^2 - d^2 + 2fh [3] [2][3]: a^2 - b^2 + 2eh = c^2 - d^2 + 2fh a^2 + d^2 - b^2 - c^2 = 2fh - 2eh [1]: 0 = 2h(f - e) Since h <>0, then f - e = 0, so e = f Last edited by Denis; July 10th, 2014 at 07:56 PM.
 October 24th, 2014, 11:57 PM #3 Newbie   Joined: Oct 2014 From: Lipowo Posts: 1 Thanks: 1 Let $R$ be a point in the intersection of the line $XO$ and the circumcircle of $WXYZ$. Inscribed angles $\angle RXY$ and $\angle WYZ$ are equal so $|RY|=|WZ|$. Therefore $WYRZ$ (or $WYZR$) is isosceles trapezoid. Thanks from aurel5

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# WXYZ is a quadrilateral whose diagonals meet at a point O such that OW=OX=OZ.angel OW X =50° ,find angle OZW

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