A chord and its perpendicular bisector

A chord and its perpendicular bisector

Secondary 4
7 marks
TGM Original Questions

A circle has centre \(O\) and radius \(13\) cm. The perpendicular from \(O\) to chord \(AB\) meets it at \(M\), where \(OM=5\) cm. Point \(P\) lies on \(MO\) produced beyond \(O\).

  1. Prove that \(AM=BM\), and find \(AB\).[4]
  2. Prove that \(PA=PB\).[3]

Solution:

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Answer:(a) \(AM=BM\); \(AB=24\) cm. (b) \(PA=PB\) by SAS.

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