July 9th, 2014, 10:13 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, 08:51 PM  #2 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 11,664 Thanks: 740 
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 08:56 PM. 
October 25th, 2014, 12:57 AM  #3 
Newbie Joined: Oct 2014 From: Lipowo Posts: 1 Thanks: 1 
Let be a point in the intersection of the line and the circumcircle of . Inscribed angles and are equal so . Therefore (or ) is isosceles trapezoid.


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geometry, homework, proof, triangles 
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