Let $ABC$ be a triangle such that $AB=55, AC=35$ and $BC=72$. Consider a line $\ell$ that intersects side $BC$ at $D$ and side $AC$ at $E$ and that divides the triangle into two figures with equal perimeters and equal areas. Determine the measure of segment $CD$.

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Let $CD = x$, $CE = y$, and $DE = z$.

(1) Since the triangle CED has the same perimeter as the quadrilateral ABDE, we have

$$
x + y + z = (35 - y) + z + (72 - x) + 55 \Longleftrightarrow y = 81 - x
$$

(2) Since they also have the same area, the area of triangle DCE must be equal to half the area of triangle $ABC$. Thus,

$$
\frac{x y \sin \hat{C}}{2} = \frac{1}{2} \cdot \frac{35 \cdot 72 \cdot \sin \hat{C}}{2} \Longleftrightarrow x y = 1260
$$

Using the two equations found, we get \( x^2 - 81x + 1260 = 0 \). Solving this equation, we obtain \( x = 60 \) or \( x = 21 \). In the first case, we get \( y = 21 \) and in the second case, \( y = 60 \). Since E is on side AC, we must have \( y \leq 35 \), so the solution we are interested in is \( x = 60 \) and \( y = 21 \). Therefore, \( CD = 60 \).

Suggestion: Calculate the area of \( \triangle CED \), which is half the area of \( \triangle ABC \).

Helpful Facts: The area \( S \) of a triangle with two sides of lengths \( a \) and \( b \) and the angle \( \theta \) between them can be calculated using the formula

$$
S = \frac{a b \sin \theta}{2}
$$

Proof: The area of the triangle in Figure 107.1 is \( \frac{ah}{2} \), but \( h = b \sin \theta \).

![](https://cdn.mathpix.com/cropped/2024_05_01_fcff7d0cdc4488e80dd9g-5.jpg?height=259&width=486&top_left_y=907&top_left_x=1436)

Figure 107.1

Thus,

$$
\frac{a h}{2} = \frac{a b \sin \theta}{2}
$$

Suggestion: Show that triangles BME and HEN are isosceles.

Helpful Facts: The orthocenter of a triangle is the point of intersection of the altitudes. In a right triangle, the median relative to the hypotenuse has a length equal to half the hypotenuse.

![](https://cdn.mathpix.com/cropped/2024_05_01_fcff7d0cdc4488e80dd9g-6.jpg?height=340&width=351&top_left_y=738&top_left_x=224)

Figure 108.1