3+ Area of a quadrilateral

In an isosceles triangle $A B C$ with an obtuse angle $A$, equal to $\alpha$, altitudes $B N$ and $C M$ are drawn.

Find the ratio of the area of quadrilateral $B M N C$ to the area of triangle $A B C$.
$S_{\text {BMNC }}=\frac{1}{2} B N \cdot C M \sin \boldsymbol{\alpha}$.

## Solution

Let $B C=a$. If $A K-$ is the height of triangle $A B C$, then

$$
A K=B K \operatorname{ctg} \angle B A K=\frac{a}{2} \operatorname{ctg} \frac{\alpha}{2}
$$

$S_{\triangle \mathrm{ABC}}=\frac{1}{2} B C \cdot A K=\frac{1}{4} a^{2} \operatorname{ctg} \frac{\alpha}{2}$.

From the right triangle $B N C$, we find that

$$
B N=B C \cos \angle N B C=a \cos \left(90^{\circ}-\right.
$$

$$
\left.\frac{\alpha}{2}\right)=a \sin \frac{\alpha}{2}
$$

Similarly, $C M=a \sin \frac{\alpha}{2}$. Then

$$
S_{\mathrm{BMNC}}=\frac{1}{2} B N \cdot C M \sin \angle B A C=\frac{1}{2}
$$

$$
a^{2} \sin ^{2} \frac{\alpha}{2} \sin \alpha
$$

Therefore,

$$
\frac{S_{B M N C}}{S_{\triangle A B C}}=\frac{2 \sin ^{2} \frac{\alpha}{2} \sin \alpha}{\operatorname{ctg}_{\frac{\alpha}{2}}}=4 \sin ^{4}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_5e9166ba6ad1f1b1f21fg-20.jpg?height=335&width=629&top_left_y=1233&top_left_x=37)

Answer

$$
4 \sin ^{4} \frac{\alpha}{2}
$$