[Ratio of areas of similar triangles Law of Sines

$A B$ and $C D$ are two non-intersecting chords, with $\cup A B=120^{\circ}$ and $\cup C D=90^{\circ} ; M$ is the point of intersection of the chords $A D$ and $B C$. Find the areas of triangles $A M B$ and $C M D$, if the sum of these areas is 100.

#
Express the sides $A B$ and $C D$ of similar triangles $A M B$ and $C M D$ in terms of the radius of the circle.

## Solution

Since

$$
\angle M A B=\angle D A B=\angle B C D=\angle M C D,
$$

triangles $A M B$ and $C M D$ are similar. If $R$ is the radius of the given circle, then

$$
A B=2 R \sin 60^{\circ}=R \sqrt{3}, C D=2 R \sin 45^{\circ}=R \sqrt{2}.
$$

Therefore, the similarity ratio is

$$
\frac{A B}{C D}=\frac{R \sqrt{3}}{R \sqrt{2}}=\frac{\sqrt{3}}{\sqrt{2}}
$$

Thus,

$$
\frac{S_{\triangle A M B}}{S_{\triangle C M D}}=\left(\frac{\sqrt{3}}{\sqrt{2}}\right)^{2}=\frac{3}{2}
$$

Therefore, $S_{\triangle \text { AMB }}=60$ and $S_{\Delta \text { CMD }}=40$.

![](https://cdn.mathpix.com/cropped/2024_05_06_5ddd4eab4fe1a12edeb9g-04.jpg?height=420&width=472&top_left_y=2114&top_left_x=793)

Answer

60 and 40.