ABC is an equilateral triangle. BP is perpendicular to AC, PR is perpendicular to AB, and PQ is perpendicular to BC.
What is the ratio of \(AB^2 : BP^2 : PR^2\)?
16 : 12 : 3
Since BP ⊥ AC in an equilateral triangle, P is the midpoint of AC and BP is an altitude, so \(BP=\frac{\sqrt3}{2}a\) where \(a\) is the side. Taking \(A=(0,0)\), \(B=(a,0)\), \(C=(a/2,\frac{\sqrt3}{2}a)\) gives \(P=(a/4,\frac{\sqrt3}{4}a)\). The foot of the perpendicular from P to AB is \(R=(a/4,0)\), so \(PR=\frac{\sqrt3}{4}a\). Hence \(AB^2:BP^2:PR^2=a^2:\frac34a^2:\frac{3}{16}a^2=16:12:3\).
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