Asked by baidshryans | 26th Nov, 2008, 06:12: PM
This problem can be solved by comparing the areas of the two triangles. The crucial idea is that angle bisectors are equally distant from both sides of the angle.
Below is a figure drawn so that it doesn't look like an isosceles
the blue lines are perpendiculars from the point D to the two
sides of the top angle. Since D is on the angle bisector, it's the
same (perpendicular) distance from the two sides BA and BC.
area of triangle ABD
A1 = (1/2)*base*height = (1/2)*a*h
where h is the altitude of the left triangle, the length of the red
line. Also area of triangle BCD
A2 = (1/2)*b*h
since the triangles have the same altitude, from the way their bases
are collinear, and have a vertex in common.
But these two areas can be figured in a different way. The triangle
on the left can be turned around so that the blue line on the left
is an altitude. Now its area is:
A1 = (1/2)*x*k
where k is the length of the blue line.
In the same way, the area of the triangle on the right is:
A2 = (1/2)*y*k
So a*h = x*k and b*h = y*k. A simple calculation shows that:
x/y = a/b
Answered by | 17th Dec, 2008, 04:51: PM
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