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 the ratio of the area of the triangle ADO to the area of the quadrilateral BDOM?
3 : 5
Using \(A=(0,h)\), \(D=(-3a/8,h/4)\), \(O=(0,h/2)\), \(B=(-a/2,0)\) and \(M=(0,0)\), the shoelace formula gives area of \(\triangle ADO=\frac{3ah}{32}\) and area of quadrilateral \(BDOM=\frac{5ah}{32}\) (their sum equals area of \(\triangle ABM=\frac{ah}{4}\), half the triangle's area, confirming consistency). Hence the ratio is \(3:5\).
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