Advanced Geometric Problem & Solution (Twice a week update) Now 4632 problems posted.
<Qua-2023-720>, hint
Let <ALE=<KCE=p, <LED=<FCE=q ( EL is tangent to circle CFE)
<FCK=<FCE+<ECK=q+P equal to <ADK=<AED=<LED+<ALE=q+p
TriaNgle ALE is isosceles AL=AE=AD.
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