Problem
GEO-B3-M06-P003 Median via Cosines
#3
★☆☆☆☆ Level 1 of 5
Let \(AM\) be a median of triangle \(ABC\). Prove that \(AB^2+AC^2=2AM^2+\frac12 BC^2\).
Inspired by Prasolov trigonometric geometry method
C. Hint 1. Denote \(BM=CM=\frac a2\).
D. Hint 2. Apply the cosine rule in \(ABM\) and \(ACM\).
Let \(\angle AMB=\theta\). Then \(\angle AMC=180^\circ-\theta\). By the cosine rule, \[ AB^2=AM^2+\frac{a^2}{4}-a\cdot AM\cos\theta, \] \[ AC^2=AM^2+\frac{a^2}{4}+a\cdot AM\cos\theta. \]
Adding gives \(AB^2+AC^2=2AM^2+\frac{a^2}{2}=2AM^2+\frac12 BC^2\).
This is Apollonius' formula, derived trigonometrically.