Question

In an acute angled triangle, express a median in terms of its sides.

Answer 1

Given in Δ ABC, AD is median.
Construction: AE ⊥ BC
Now since AD is the median
∴ BD = CD = 1/2 BC ....(1)
In Δ AED 
AD² = AE² + DE² (Pythagoras Theorem)
⇒ AE² = AD² - DE² .....(2)
In Δ AEB
AB² = AE² + BE²
⇒ AD² - DE² + BE² {From (2)}
= (BD + DE)² + AD² - DE² (∴ BE = BD + DE)
BD² + DE² + 2BD*DE + AD² - DE²
= BD² + AD² + 2BD*DE
= (1/2BC)² + AD² + (2×1/2BC×DE)   {From (1)}
= (1/4BC)² + AD² + BC*DE ....(3)
In Δ AED
AC² = AE² + EC²
= AD² - DE² + EC²
= AD² - DE² + (DC - DE)²
= AD² - DE² + DC² + DE² - 2DC*DE
AD² + DC² - 2DC*DE
= AD² + (1/2BC)² - (2×1/2BC*DE)
= AD² + (1/4BC)² - BC*DE ....(4)
Adding (3) and (4), we get 
AB² + AC² =1/4BC² + AD² + BC*DE + AD² + 1/4BC² - BC*DE 
= 1/2BC² + 2AD²
2(AB² + AC²) = BC² + 4AD²
2AB² + 2AC² = BC² + 4AD²  Answer.

Answer 2

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