A, B and C are points on the circumference of a circle. Angle CAB = angle CBA. DA and DB are tangents to the circle from the point D. Prove that triangle ACD and triangle BCD are congruent. Answer

1. Use the equal angles in triangle $ABC$ to identify an equal pair of sides.

In $\triangle ABC$,
$\angle CAB = \angle CBA.$
Hence the sides opposite these equal angles are equal:
$BC = AC.$

2. Use the tangent property to identify another equal pair of sides.

Since $DA$ and $DB$ are tangents to the circle from the same external point $D$,
$DA = DB.$

3. State the common side and conclude congruence.

The side $CD$ is common to both triangles, so
$CD = CD.$
Thus, in $\triangle ACD$ and $\triangle BCD$:
$AC = BC, \quad AD = BD, \quad CD = CD.$
Therefore,
$\triangle ACD \cong \triangle BCD \quad \text{(SSS)}.$