Advanced Geometric Problem & Solution (Twice a week update) Now 4666 problems posted.
<Tri-2025-2254>, hint
<BDM=<ACD=X, <BCD=Y
Consider line MD cuts line CA at A'
Since <ADA'=<BDM=X =<ACD and <BAC=X+Y=<BCA, <AA'M=Y=<DMB, M on S" Hence BM//AC
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