Geometry Proof Similarity

Question

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.

Answer 1

Triangles PWZ and OWX are similar; PYZ and OXY are similar, because both triangles are right-angled and angles PWZ and OXW are equal; and angles OXY and PYZ are equal (complementary angles).

OW/OX=PZ/PW; OY/OX=PZ/PY. OW=OX.PZ/PW and PY=OX.PZ/OY.

Therefore, OW/PY=OY/PW=(OP+PY)/(OP+OW)

Cross-multiplying: OW.OP+OW^2=OP.PY+PY^2

Rearranging: OP(OW-PY)+OW^2-PY^2=0=OP(OW-PY)+(OW+PY)(OW-PY)

Thus: (OW+OP+PY)(OW-PY)=0=WY(OW-PY)=0

Since the diagonal WY is non-zero, OW=PY (QED).