Given: In ∆ABC, D is mid-point of BC since AD is median AM ⊥ BC and AC > AB.

To prove:

(i) AC² = AD² + BC.DM + \({ \left( \frac { BC }{ 2 } \right) }^{ 2 }\)

(ii) AB² = AD² – BC.DM + \({ \left( \frac { BC }{ 2 } \right) }^{ 2 }\)

(iii) AC² + AB² = 2AD² + \(\frac { 1 }{ 2 }\) BC²

Proof : (i) In right angled ∆AMD

AD² = AM² + DM²

AM² = AD² – DM²

in right ∆AMC

AC² = AM² + MC² …..(ii)

From equation (i) and (ii),

AC² = (AD² – DM²) + MC²

⇒ AC² = (AD² – DM²) + (DM + DC)² (∵ MC = DM + DC)

⇒ AC² = AD² – DM² + DM² + DC² + 2DM.DC

AC² = AD² + DC² + 2DM.DC

AC² = AD² + \({ \left( \frac { BC }{ 2 } \right) }^{ 2 }\) + 2DM·\(\frac { BC }{ 2 }\)

Thus, AC² = AD² + BC.DM + \({ \left( \frac { BC }{ 2 } \right) }^{ 2 }\) …..(iii)

(ii) In right ∆AMB

AB² = AM² + BM²

= (AD² – DM²) + BM² [using equation (i)]

= (AD² – DM²) + (BD – DM)²

= AD² – DM² + BD² + DM² – 2 BD.DM

AD² + BD² – 2BD.DM

= AD² + \({ \left( \frac { BC }{ 2 } \right) }^{ 2 }\) – 2 × \(\frac { 1 }{ 2 }\) BC. DM

∴ AB² = AD – BC.DM + \({ \left( \frac { BC }{ 2 } \right) }^{ 2 }\)

Thus, AB² = AD – BC.DM + \({ \left( \frac { BC }{ 2 } \right) }^{ 2 }\) ……(iv)

Adding equation (iii) and (iv)

AB² + AC² = 2AD² + 2 × \(\frac { 1 }{ 4 }\) BC²

= 2AD² + \(\frac { 1 }{ 2 }\) BC²

Thus, AB² + AC² = 2AD² + \(\frac { 1 }{ 2 }\) BC²