The CSAT and Practice[+] are designed by the Climb Foundation to help candidates. We are advocates for more opportunity to shine and less opportunity to fail, and we strive to level the playing field. Practice Paper 1 Question 3

Triangle $$ABC$$ is isosceles with $$AB=AC$$. Let the circle having diameter $$AB$$ and centre $$O$$ intersect $$BC$$ at some point $$P$$. Find the ratio $$\frac{BP}{BC}$$.

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Hints

• Hint 1
What can you say about the $$\angle APB$$?
• Hint 2
What about $$AP$$?

Solution

Angle $$\angle APB=\pi/2$$ because it subscribes an arc length of $$\pi$$ (AB is a diameter), hence $$AP$$ is the height in the triangle. Being an isosceles triangle, this is also the median, and hence $$\frac{BP}{BC}=\frac{1}{2}$$. Can you find an alternative proof considering the segment $$OP$$ instead?

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