Let $P$ be the intersection point of line $OD$ and the line passing through point $A$ parallel to $OM$. Then triangle $AOP$ is isosceles.

## Solution

Assume that point $C$ is located between points $O$ and $D$. Draw a line through point $A$ parallel to $MO$. Let $P$, $Q$, and $T$ be the points of intersection of this line with line $OD$, the angle bisector of $\angle AOD$, and line $MB$ respectively.

Triangle $POA$ is isosceles because its angle bisector $OQ$ is also the altitude. Therefore, $OP = OA = 12$. Hence, $PD = OP - OD = 4$.

From the similarity of triangles $MDO$ and $ADP$, we have $MO = AP \cdot OD / DP = 2AP$.

From the similarity of triangles $PCT$ and $OCM$, we find: $PT = MO \cdot PC / CO = MO = 2AP$. Therefore, $AT = AP$ and $MO = 2AT$.

From the similarity of triangles $MBO$ and $TBA$, we get $OB: AB = MO: AT = 2: 1$.

If $S_{AOD} = S$, then $S_{BOC} = 2/3 \cdot 6/8 S = S/2$. Therefore, $S_{ABCD} = S_{AOD} - S_{BOC} = S/2$.

If point $D$ is located between points $O$ and $C$, then by similar reasoning, we find that $S_{AOD}: S_{ABCD} = 14$:

11. 

![](https://cdn.mathpix.com/cropped/2024_05_06_f7f9d90261bfc416bd71g-39.jpg?height=368&width=829&top_left_y=845&top_left_x=617)

## Answer

$2: 1$ or $14: 11$.
8,9 |  |

The extensions of sides $A D$ and $B C$ of the convex quadrilateral $A B C D$ intersect at point $M$, and the extensions of sides $A B$ and $C D$ intersect at point $O$. The segment $M O$ is perpendicular to the bisector of angle $A O D$. Find the ratio of the areas of triangle $A O D$ and quadrilateral $A B C D$, if $O A=12, O D=8, C D=2$.