# Elementary geometry question involving quadrilateral and bisector given with picture.

• $ABCD$ 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 $\color{magenta}{24^\circ}$.

• I agree; I think there are two degrees of freedom still. Are you sure there weren't other assumptions in the problem, like line $BD$ bisecting angle $D$ also, or $ABCD$ being a special type of quadrilateral? – Greg Martin Aug 29 '15 at 22:53
• @GregMartin, yes, above are the only given things. I checked that $x=24^\circ$ no matter what $\widehat{CBA}$ values; so the question has an unique solution. – Alistair Aug 30 '15 at 1:00

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, $x = \angle CAX = \dfrac {48}{2} = 24^0$
• @GregMartin If a quadrilateral is cyclic, the angles in the same segment should be equal. $\angle CAX$ and $x$ are instances of such. $\angle ADB$ and $p$ formed another equal pair. Remember also that AX is an angle bisector of $\angle BAC$ – Mick Aug 31 '15 at 6:14