Advanced Geometric Problem & Solution (Twice a week update) Now 4503 problems posted.
<Tri-2024-2138>, hint
1. Triangles BAC and MCD are similar.
BD and ME are midlines. Hence triangles CME and ABD are similar.
2.Since <ABD=<CME. <BEM=90=<CMB+<ABD=<CME+<CMB=90.
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