In triangle ABC, CD is the angular bisector, CF is the outer angular bisector, DF is parallel, BC intersects AC and E, CF intersects F, and de = EF is proved

Prove: take a point G on the extension line of BC, because DF / / BC, so angle FCG = angle DFC, because CF bisects the outer angle, so angle FCG = angle ACF, so angle DFC = angle ACF, so EF = EC, because DF / / BC, so angle CDF = angle BCD, and because CD bisects the outer angle BCA, so angle FCG = angle ACF, so angle DFC = angle ACF, so EF = EC

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