Question

在△ABC中,∠C=70°,AD是△CAB的平分线,BD是△ABC的外角平分线,AD于BD交于点D,求∠D的度数

Answer 1

∠D的度数为:70/2=35°.

设,∠CAD=∠DAB=∠1,

∠CBD=∠DBE=∠2.

∠ABC=180-(∠C+2∠1),

而,∠ABC=180-2∠2,

则有∠C+2∠1=2∠2,

∠2-∠1=∠C/2=35°.

∠D=180°-(∠1+∠ABC+∠2)

=180-∠1-∠2-(180-2∠2)

=∠2-∠1

=35°