The diagonals $AC$ and $BD$ of a convex quadrilateral $ABCD$ intersect in point $M$. The angle bisector of $\angle ACD$ intersects the ray $\overrightarrow{BA}$ in point $K$. If
$MA.MC+MA.CD=MB.MD$,
prove that $\angle BKC=\angle CDB$.
Well know problem, Let $E$ an point in $AC$ such that $CD=CE$. For the condition $MA.MC+MA.CD=MB.MD=MA.ME$ So $ABED$ is a cyclic cuadrilateral. Now is Angle Chasing $\angle ACK=\angle KCD=\angle AED$ For the $ABED$ cyclic $\angle AED=\angle ABD$, So $KBCD$ is cyclic. Now $\angle BKC=\angle CDB$.
Mathsy123 wrote:
Well know problem, Let $E$ an point in $AC$ such that $CD=CE$. For the condition $MA.MC+MA.CD=MB.MD=MA.ME$ So $ABED$ is a cyclic cuadrilateral. Now is Angle Chasing $\angle ACK=\angle KCD=\angle AED$ For the $ABED$ cyclic $\angle AED=\angle ABD$, So $KBCD$ is cyclic. Now $\angle BKC=\angle CDB$.
$MA(2.CE+ME)=MA.ME$? I don't understand that part.