Elementary geometry question involving quadrilateral and bisector given with picture.

Question

• $ABCD$ is a quadrilateral.
• $m(\widehat{BAC})=48^\circ$.
• $m(\widehat{CAD})=66^\circ$.
• $m(\widehat{CBD})=m(\widehat{DBA})$.
• What is $\color{magenta}{m(\widehat{BDC})=x}$?

Tried lots of things but all i can found is that all the angles, except given ones and $\color{magenta}x$, depends on the $m(\widehat{CBA})$. The answer is $\color{magenta}{24^\circ}$.

Answer 1

Construction: Let the angle bisector of $\angle BAC$ cut $BD$ at $X$. Then, Join $CX$.

$X$ is actually is the in-center of $\triangle ABC$. Thus, $p = q$

$r = t + q = \dfrac {2t + 2q }{2} = \dfrac {180 - 48}{2} = 66^ 0 = \angle CAD$

This means $XADC$ is a cyclic quadrilateral

Therefore, by angle in the same segment, $x = \angle CAX = \dfrac {48}{2} = 24^0$