Mi sapete risolvere questo problema geometrico?

Considero I triangoli AMC e BMD essi hanno 1) BM= MC per ipotesi. 2) AM=MD " " 3) AM^C= BM^D perché opposti al vertice. Sono verificate le ipotesi del primo criterio di congruenza quindi I triangoli AMC =BMD e valgono le seguenti relazioni: 4)BD^M=MA^C 5) AC^M=DB^M 6) AC=BD.

Considero I triangoli ABC e BDC essi hanno 1) BC in comune 2) AC=BD per la dimostrazione precedente ( punto 6) 3) AC^B=CB^D × dimostaz prec. (5) sono verificate le ipotesi 1°C .C. Allora ABC=BDC in paricolare gli angoli CA^B=CD^B.

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