Let I midpoint of $AB$ and $J =CI\cap OK$ ; we have Area$(CBI)$=Area$(CAI)$ besides $OI \parallel CK \implies $ Area $(COI)= $ Area $(KOI) \implies $ Area $(COJ)=$ Area $(JKI)$ hence the result follows $\square$

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Let $M$ be a midpoint of $AB$. $CK,OM$ are both perpendicular to $AB$, hence $[CKO]=[CKM]$Then $\frac{1}{2}[ABC]=[ACM]=[ACK]+[CKM]=[ACK]+[CKO]=[ACOK]$. Now $[BCOK]=[ABC]-[ACOK]=\frac{1}{2}[ABC]=[ACOK]$

Stronger. Let $ABC$ be a traingle such that $90^\circ\ge\angle ABC\wedge 90^\circ\ge\angle BAC$ . Point $K$ is a foot of altitude through vertex $C$. Point $O$ is given point on bisector of segment $AB$ inside triangle $ABC$. Prove that areas of quadrilaterals $AKOC$ and $BKOC$ are equal.

It seems as though I couldn't think of doing this synthetically, and in my diagram, $AKOC$ is concave, Anyways, here's a bash, Let $OC \cap AB= L$ Of course, $$ [BOC]=\frac{\sin 2A}{\sum \sin 2A} , [AOC]=\frac{\sin 2B}{\sum \sin 2A} , [KOB]= \left( \frac{\sin 2C}{\sum \sin 2A} \right) \left( \frac{\tan A}{\tan A+\tan B} \right) \text{ and } [AOK]=\left( \frac{\sin 2C}{\sum \sin 2A} \right) \left( \frac{\tan B}{\tan A+\tan B} \right) $$Now, $[BOC]+[KOB]=[AOC]+[AOK]$ Therefore, $$\frac{\sin 2A-\sin 2B}{\sum \sin 2A}=\frac{\sin 2C}{\sum \sin 2A} \left( \frac{\tan B-\tan A}{\tan B+\tan A} \right) $$$$\cos (A+B)\sin(A-B) =\sin C \cos C \left( \frac{\tan B- \tan A}{\tan B+\tan A} \right) $$$$\sin (A-B)= \sin C \left( \frac{\sin A \cos B - \sin B \cos A}{\sin A \cos B + \sin B \cos A} \right) \implies \sin (A+B)= \sin C \text{ which is indeed true,}$$Hence, $[BKOC]=[AKOC]$ is proved