Ra, Rb = orthogonal projections of R on CA, CB respectively Qa, Qb = orthogonal projections of Q on RRa, RRb respectively Pa, Pb = orthogonal projections of P on RRa, RRb respectively O' = orthogonal projections of O on CR
POQ is isosceles triangle because angle(OQP) = angle(OPQ) = 90 - angle(ACB)/2 so PO' = QO', moreover O' is midpoint of chord CR, therefore QC = PR. It makes two right triangles CLQ = PPaR and QL = RPa. As constructed QL//RPa therefore QLPaR is parallelogram. Similarly PKQbR is also parallelogram.
Denote (ABC) = area of ABC (QLR) = (QPaR) = (QQaR) - (QQaPa) (PKR) = (PQbR) = (QQbR) - (QPQb) = (QQbR) - (QQbPb) By symmetry of angle bisector CR, these triangles are conguent and we are done.
1. "RE: Triangles With Equal Area"
In response to message #0
Dear Alex,
Please note that in the proof we use important fact that P, Q are reflections each other in midpoint of CR. So the fact can be generalized as: P, Q are any two points on CR and they are reflections each other in midpoint of CR. The area of triangle bonded by R, this point and its orthogonal projection on one side AC or BC is equal area of triangle bonded by R, other point and its orthogonal projection on other side. The proof is the same.
4. "RE: Triangles With Equal Area"
In response to message #1
Dear Alex,
Here is another variant of the proof (may be with more nice and symmetrical picture. Please see attach file): Suppose AC<=BC and P, Q are reflections each other in midpoint of CR. A line through O and perpendicular with CR intersects CR at O', AC at D, BC at E. CDRE is a rhombus with center O'. A line through Q and perpendicular with BC (RD) intersects BC at L1, RD at Q' A line through P and perpendicular with AC (RE) intersects AC at K1, RE at P' By symmetries of rhombus in center O': (RPK) = (RQL1K) - (PQL1K) = (RQLK1) - (KL1Q'K2)/2 = (RQLK1) - (LK1P'L2)/2 = (RQLK1) - (LK1R) = (RQL)
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