In the diagram, $D$ is on side $A C$ of $\triangle A B C$ so that $B D$ is perpendicular to $A C$. If $A B=29$, $A C=69$, and $B D=20$, what is the length of $B C$ ?

![](https://cdn.mathpix.com/cropped/2024_04_17_24bfa07284b5df794287g-3.jpg?height=384&width=1065&top_left_y=735&top_left_x=519)

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By the Pythagorean theorem in $\triangle A B D, A D^{2}+20^{2}=29^{2}$, so $A D^{2}=841-400=441$.

Since $A D>0, A D=\sqrt{441}=21$.

This means $D C=A C-A D=69-21=48$.

Notice that $5^{2}+12^{2}=25+144=169=13^{2}$. Therefore, $16\left(5^{2}\right)+16(12)^{2}=16(13)^{2}$ or $(4 \times 5)^{2}+(4 \times 12)^{2}=(4 \times 13)^{2}$, so $20^{2}+48^{2}=52^{2}$.

Applying the Pythagorean theorem to $\triangle C D B$, we have

$$
C B^{2}=B D^{2}+C D^{2}=20^{2}+48^{2}=52^{2} \text {. }
$$

Since $C B>0, C B=52$.