Area of triangle DBC: Can you solve it? #maths

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Area of triangle DBC: Can you solve it? #maths

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Cool. My firt thought was to say BD = sqrt(27) and go from there, but BD can't be perpendicular to AC, so that was out of the window.

walter
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BD^2 = h^2+9
BC^2 = h^2+144
take reflection of triangle BAD in line BA, we get a new point E on line CA
by angle bisector theorem
BE/BC = 6/9
(h^2+9)/(h^2+144) = 4/9 ( because BE = BD)
5 h^2 = 495
h = 3√11
area of triangle BDC = (1/2)(9)(3√11) = 27√11/2

raghvendrasingh
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