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15. Two circles intersect at points A and B. From a pointon the common chord BA produced, secants PCDand PEH are drawn one to each circle. Prove that thepoints C, D, H, E are concyclic. |
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Answer» The quadrilateral CDHE is required to be proved to be a cyclic quadrilateral. CDH+CEH=180=DCE+EHD. Join CB and AD, and join AH and BE. From this construction we get two pairs of similar triangles: APD and BPC, and APH and BPE, because of the common angle at P, and equal angles PCB=DAP, PAH=BEP (angles in the same segment). We can therefore write: PD/PB=PA/PC=DA/BC (triangles PCB and DAP) and PB/PH=PE/PA=BE/HA (triangles PAH and BEP). From this we get: PB.PA=PC.PD=PE.PH. So PC.PD=PE.PH PC/PE=PH/PD. Therefore, triangles PCE and PHD are also similar because P is theincluded common angle. PDH=PEC. But CDH=180-PDH (supplementary angles on a straight line), so CDH=180-PEC (PEC is the same angle as CEH). These are opposite angles of the quadrilateral CDHE, and this is a definitive property of cyclic quadrilaterals, so CDHE is cyclic. |
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