Prove Apollonius theorem: In \(\Delta ABC\), if M is the midpoint of side BC, then \(AB^2 + AC^2 = 2AM^2 + 2BM^2\).

Answer: Proof of Apollonius theorem

[Figure: \(\Delta ABC\) with midpoint M of BC and seg AD \(\perp\) BC. The figure is drawn with D between M and C; the other cases are proved the same way.]

Given: M is the midpoint of BC, so BM = MC. Construction: Draw seg AD \(\perp\) BC.

In \(\Delta AMC\), \(\angle AMC\) is acute and D lies on MC. In right-angled \(\Delta ADC\) and \(\Delta ADM\): \(AC^2 = AD^2 + DC^2 = AD^2 + (MC - MD)^2\) and \(AD^2 = AM^2 - MD^2\).

\(\therefore AC^2 = AM^2 - MD^2 + MC^2 - 2\,MC \times MD + MD^2 = AM^2 + MC^2 - 2\,MC \times MD\) ...... (I)

In \(\Delta AMB\), \(\angle AMB\) is obtuse and B – M – D. Similarly \(AB^2 = AD^2 + BD^2 = AM^2 - MD^2 + (BM + MD)^2 = AM^2 + BM^2 + 2\,BM \times MD\) ...... (II)

Adding (I) and (II) and using BM = MC: \(AB^2 + AC^2 = 2AM^2 + BM^2 + MC^2 + 2\,BM \times MD - 2\,MC \times MD = 2AM^2 + 2BM^2\). Hence proved.