Triangle with Special Point D

Geometry Level 2

In triangle $ABC$ , $\angle ABC = 2 \alpha$ and $\angle ACB = 140 ^\circ - \alpha$ . $D$ is a point on line segment $AB$ such that $CB = BD$ . What is the measure (in degrees) of $\angle DCA$ ?

55 40 50 45

1 solution

Arron Kau Staff
May 13, 2014

Since $BC = BD$ , hence $\angle BCD = \angle BDC = 90 ^ \circ - \alpha$ .

Thus $\angle DCA = \angle BCA - \angle BCD = (140^ \circ - \alpha) - ( 90^\circ - \alpha) = 50 ^\circ$ .

Triangle with Special Point D

1 solution

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