A circle with centre O passes through the vertex A of an equilateral triangle ABC and touches BC at its midpoint M. The circle cuts AB at D and AC at E.
What is \(AD : DB\) equal to?
3 : 1
Let the side of equilateral triangle ABC be \(a\). Take \(M\) (midpoint of BC) as origin, with \(A=(0,h)\), \(B=(-a/2,0)\), where \(h=\frac{\sqrt3}{2}a\). Since the circle touches BC at M, its centre O lies on line AM (the perpendicular to BC at M). As \(OA=OM=\) radius, O must be the midpoint of AM, so \(O=(0,h/2)\) with radius \(h/2\). Substituting the parametric point on AB, \((-ta/2,\,h(1-t))\), into the circle's equation \(x^2+(y-h/2)^2=(h/2)^2\) gives roots \(t=0\) (point A) and \(t=3/4\) (point D). So \(AD=\frac34AB\) and \(DB=\frac14AB\), giving \(AD:DB=3:1\).
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