We have triangle ABC where AB = AC and AD is an altitude. Meanwhile, E is a point on AC such that AB is parallel to DE. If BC = 15 and the area of triangle ABC is 200, what is the area of ABDE?
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E
B D C
If AB = AC then the triangle is isosceles
If BC is 15 then AD = 2 * 200 / 15 = 80/3
And because ABC is isosceles, then DC = BD = 15 /2
And since DE is parallel to AB and AC is a transversal cutting both, then triangles ABC and EDC are similar because angle ACD = angle ECD and angle BAC = angle DEC
And since BC = 2 DC.. then 2 is the scale factor between triagles ABC and EDC.....then the area of triangle ABC = area of triangle EDC * scale factor ^2
So
200 = area of EDC * 2^2
200 = area of EDC * 4
200 / 4 = area of EDC = 50
So [ ABDE ] = area of ABC - area of EDC = 200 - 50 = 150
