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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\$?

Aug 16, 2018

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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

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   Aug 16, 2018
edited by CPhill  Aug 16, 2018