Find the distance BC, quarter circle
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Solution
We have a right angle at the point \(D\), thus the line \(CD\) can be extended to a point \(E\) such that \(AE\) is a diameter of the circle like it is shown in the next figure
Now, let’s connect the two points \(A\) and \(C\) and focus on the triangle \(\triangle ADC\)
Using Pythagorean theorem
\[\begin{aligned}
A C^{2} &=A D^{2}+D C^{2} \\\\
A C^{2} &=24^{2}+7^{2} \\\\
&=576+49 \\\\
&=625 \\\\
A C &=25
\end{aligned}\]
The triangles \(\triangle ABC\) and \(\triangle EBC\) are congruent triangles, therefore \[CE=AC=25\]
\[\begin{aligned}
&D E=D C+C E \\\\
&D E=7+25 \\\\
&D E=32
\end{aligned}\]
Let’s focus on the triangle \(\triangle ADE\) (Right triangle)
\[\begin{aligned}
A E^{2} &=A D^{2}+D E^{2} \\\\
&=24^{2}+32^{2} \\\\
&=576+1024 \\\\
&=1600 \\\\
A E &=40
\end{aligned}\]
Let’s focus on the triangle \(\triangle BEC\) (Right triangle)
\[\begin{aligned}
C E^{2} &=B C^{2}+B E^{2} \\\\
B C^{2} &=C E^{2}-B E^{2} \\\\
&=25^{2}-20^{2} \\\\
&=625-400 \\\\
&=225 \\\\
B C &=15
\end{aligned}\]
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