Two circles meet at E and F.
A line meets the circles at points P,Q, R and S in that order.
Find : ∠PEQ/∠RFS
Let A be the intersection of line PQRS
and the common chord EF (excluding the
points E and F).
Equal Inscribed Angles:
/APE = /RPE = /RFE = /RFA (1)
/ASF = /QSF = /QEF = /QEA (2)
Vertical Angles:
/EAQ = /EAP = /FAS = /FAR (3)
Similar Triangles:
(1)&(3) ==> ^EAP ~ ^RAF (4)
(2)&(3) ==> ^FAS ~ ^QAE (5)
Proof:
(4) ==> /PEA = /FRA
/PEA = /PEQ + /QEA (6)
/FRA an external angle of ^FRS
==> /FRA = /SFR + /RSF
= /RFS + /QSF (7)
Combining (2), (6), and (7) gives
/PEQ = /RFS
Therefore, the ratio is one.
QED
Note: /XYZ denotes angle XYZ and
^XYZ denotes triangle XYZ.
Edited on May 22, 2016, 8:21 pm
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Posted by Bractals
on 2016-05-22 20:19:56 |
Solution
