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Geometry Difficulty 5.0 AIME Prove it Mongolia

A circle touches BCBC side of triangle ABCABC at the point MM and intersects sides ABAB and ACAC at DD and EE respectively. If DEBCDE \parallel BC, prove that EM=MDEM = MD.

Solutions — 2

Solution 1

Since BCBC is tangent to the circle, we have EMC=EAM=EDM\angle EMC = \angle EAM = \angle EDM and DMB=DAM=DEM\angle DMB = \angle DAM = \angle DEM. From DEBCDE \parallel BC, we have EDM=DMB\angle EDM = \angle DMB. So AMAM bisects CAB\angle CAB.

Solution 2

Since BCBC is tangent to the circle, we have EMC=EAM=EDM\angle EMC = \angle EAM = \angle EDM and DMB=DAM=DEM\angle DMB = \angle DAM = \angle DEM. From DEBCDE \parallel BC, we have EDM=DMB\angle EDM = \angle DMB. So AMAM bisects CAB\angle CAB.

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