4. Let triangle $ABC$ be an isosceles triangle with base $\overline{AB}$ of length $10 \, \text{cm}$ and legs of length $13 \, \text{cm}$. Let $D$ be a point on side $\overline{BC}$ such that $|BD|:|DC|=1:2$ and let $E$ be a point on side $\overline{CA}$ such that $|CE|:|EA|=1:2$. Calculate $|DE|$.
First method:

Let $P$ be the midpoint of the base $\overline{A B}$. Since triangle $A B C$ is isosceles, $\overline{C P}$ is perpendicular to $\overline{A B}$. Draw perpendiculars from points $D$ and $E$ to $\overline{C P}$ and mark the feet of these perpendiculars as points $F$ and $G$, respectively. Extend the line segment $\overline{F D}$ beyond point $F$ and mark a point $H$ on this line such that $H F G E$ is a rectangle. Then $\triangle H D E$ is a right triangle with hypotenuse $\overline{D E}$.

![](https://cdn.mathpix.com/cropped/2024_05_30_46c615759f04f426c464g-4.jpg?height=762&width=642&top_left_y=384&top_left_x=707)

Sketch:

1 POINT

(Note: If the ray $\overrightarrow{D F}$ (extension of line segment $\overline{D F}$) and the perpendicular from point $E$ to line segment $\overline{D F}$ are not drawn on the sketch, do not award 1 POINT.)

From the conditions that points $E$ and $D$ satisfy, $|C E|:|E A|=1: 2$ and $|B D|:|D C|=1: 2$, it follows that $|C E|:|C A|=1: 3$ and $|D C|:|B C|=2: 3$ (i.e., $|C E|=\frac{13}{3} \mathrm{~cm}$ and $|D C|=\frac{26}{3} \mathrm{~cm}$) and $|A P|=5 \mathrm{~cm}$.

From the right triangle $A P C$, it follows that $|P C|=\sqrt{|A C|^{2}-|A P|^{2}}=12 \mathrm{~cm}$.

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Triangles $E G C$ and $A P C$ are similar (by the AA similarity criterion),

1 POINT

so $|E G|:|A P|=|C E|:|C A|=1: 3$, i.e., $|E G|=\frac{|A P|}{3}=\frac{5}{3} \mathrm{~cm}$

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and $|G C|:|P C|=1: 3$, i.e., $|G C|=\frac{|P C|}{3}=4 \mathrm{~cm}$.

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Triangles $D F C$ and $P B C$ are similar (by the AA similarity criterion),

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so $|F D|:|P B|=|D C|:|B C|=2: 3$, i.e., $|F D|=\frac{2|P B|}{3}=\frac{10}{3} \mathrm{~cm}$

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and $|F C|:|P C|=2: 3$, i.e., $|F C|=\frac{2|P C|}{3}=8 \mathrm{~cm}$.

1 POINT

By applying the Pythagorean theorem to triangle $E H D$, we get

$|D E|=\sqrt{|H D|^{2}+|E H|^{2}}=$

$=\sqrt{(|H F|+|F D|)^{2}+|F G|^{2}}=$

$=\sqrt{(|E G|+|F D|)^{2}+(|F C|-|G C|)^{2}}=$

$=\sqrt{5^{2}+4^{2}}=\sqrt{41} \mathrm{~cm}$.