The Computer Laboratory

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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