+15 We have triangle ABC$ where $AB = AC$ and $AD$ is an altitude. Meanwhile, $E$ is a point on $AC$ such that $AB \parallel DE.$ If $BC = 12$ and the area of $\triangle ABC$ is $180,$ what is the area of $ABDE$?
+128407 Triangle ABC is isosceles with AB = AC
AD is an altitude which is also a median to BC...so DC = DB = 6
And DE is parallel to AB
So angle EDC = angle ABC
And angle ACB = angle ECD
So...by AA congruency, ΔABC ~ Δ EDC
Since the area of Δ ABC = 180....we can find the altitude AD thusly
180 = (1/2) (AC) (AD)
180 = (1/2) (12) (AD)
180 = 6 (AD) divide each side by 6
30 = AD
Now Δ ABC and Δ EDC are similar.... and BC = 2DC
So......the altitude of Δ ABC is also twice that of Δ EDC...so the altitude of Δ EDC = 15
And the area of Δ EDC = (1/2)DC( altitude of EDC) = (1/2)(6)(15) = .45
So...the area of ABDE = area of Δ ABC - area of Δ EDC = 180 - 45 = 135 units^2
