[The ratio of the areas of triangles with a common base or common height] Complex

In triangle $ABC$, altitudes $AD$ and $CE$ are drawn. Find the ratio of the areas of triangles $ABC$ and $AED$, if $AB=6$, $AC=5$, and $CB=7$.
Using the Pythagorean theorem, set up equations for $A E$ and $B D$.

## Solution

Let $A E=x, B D=y$. Then

$$
B E=6-x, C D=7-y, 25-x^{2}=49-(6-x)^{2}, 36-y^{2}=25-(7-y)^{2}
$$

From these equations, we find that $x=1, y=\frac{30}{7}$.

Let $D P$ be the height of triangle $A D E$. From the similarity of triangles $B P D$ and $B E C$, it follows that

$$
D P=\frac{E C \cdot B D}{B C}=\frac{30}{49} E C
$$

Therefore,

$$
\frac{S_{\triangle A B C}}{S_{\triangle A E D}}=\frac{A B \cdot E C}{A E \cdot D P}=\frac{A B \cdot E C}{\frac{30}{49} E C}=6 \cdot \frac{49}{30}=\frac{49}{5}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_b9b3326e3456e69a696eg-45.jpg?height=360&width=656&top_left_y=2504&top_left_x=701)

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