In the given triangle, CD is the bisector of ∠BCA. CD = DA. If ∠BDC = 76°, what is the degree measure of ∠CBD?

66°

 Concept used

The two angles of an isosceles triangle, opposite to equal sides are equal in measure.

Angle sum property = Sum of all the three angles of a triangle is \({180}^\circ\).

Calculations

In \(\Delta \) ABC

CD is the bisector of ∠BCA

⇒ ∠BCA = ∠DCA = \(\Theta \)

Since, CD = DA

⇒ ∠DCA = ∠CAD = \(\Theta \)

⇒ ∠BDC = \({76}^\circ\)

⇒ ∠BDC = ∠DCA + ∠CAD

⇒ \(\Theta \) + \(\Theta \) = \({76}^\circ\)

⇒ 2\(\Theta \) = \({76}^\circ\)

⇒ \(\Theta \) = \({38}^\circ\)

In \(\Delta \)CBD,

∠BCD + ∠CDB + ∠CBD = \({180}^\circ\)

⇒ \(\Theta \) + \({76}^\circ\) + ∠CDB = \({180}^\circ\)

⇒ \({38}^\circ\) + \({76}^\circ\) + ∠CDB = \({180}^\circ\)

⇒ ∠CDB = \({180}^\circ\) - \({114}^\circ\)

⇒ ∠CDB = \({66}^\circ\)

Therefore, answer is \({66}^\circ\)