See complete Problem 74
Three Intersecting Circles, Cyclic Quadrilateral, Angles. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
Monday, May 19, 2008
Elearn Geometry Problem 74
Labels:
angle,
circle,
cyclic quadrilateral
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Ang MBE = ang EAM = α ( have same arc EFM )
ReplyDeleteang EAH = ang EDH ( have same arc EH, AEHD concyclic )
α = ß
[Here I am using a for alpha and b for beta]
ReplyDeleteFrom Problem 73, we know that AEHD is a cyclic quad
So <EAH=<EDH=b (< in same seg.)
Consider triangle AEB
b+<AEB+<MAB+<EBA=180
b=180-<AEB-<MAB-<EBA-------(1)
Consider triangle MAB
<MAB+<AMB+<EBA+a=180
a=180-<AMB-<MAB-<EBA---------(2)
Since <AEB=<ABM (< in same seg)
So (1)=(2)