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# altitudes in triangles

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In triangle $ABC$, we have that $AB = AC,$ and $\overline{AD}$ is an altitude.   Also, $E$ is a point on $\overline{AC}$ such that $\overline{BE} \perp \overline{AC}.$ If $BC = 12$ and the area of triangle $ABC$ is $180,$ what is the area of $ABDE$?

Sep 7, 2023

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A

E

B        D             C

12

Since  AB = AC, the triangle is isoceles

AB bisects  BC  so that DC  = 6

And

[ABC ] = (1/2) (BC) (AD)

180  = (1/2) ( BC) (AD)

180 / 6 = AD =  30

And angle ADC  = angle BEC  = 90°

And angle DCA  = angle ECB

So  triangle CDA   is similar to triangle CEB

So   CD  / AD   = CE / BE

So    6/30  =  CE / BE

So     1/5  = CE / BE

5CE  = BE

And

BC^2  =  CE^2 + BE^2

12^2  =   CE^2 + (5CE)^2

144 =      CE^2 + 25CE^2

144  =   26CE^2

144/26 = CE^2

72/13 = CE^2

CE =  sqrt [ 72/13] =  (6 sqrt (2)) / sqrt (13)

tan (angle ACD)  = AD / DC  =  30/6 =  5/1

sin (ACD)  = sin (BCE) = 5  /sqrt (5^2 + 1^2)  =  5/sqrt (26)

Area of triangle  ECD  = (1/2)(DC) (CE)  = (1/2)(6)(6sqrt (2)  /sqrt (13) )  * (5 /sqrt (26))   =

90sqrt (2) / sqrt (13) / [ sqrt (2) * sqrt (13) ]   = 90  / 13

[ ABDE ]  = [ ABC ] - [ ECD ]  =   (180 )  - ( 90/13 )    = 2250 / 13   Sep 7, 2023
edited by CPhill  Sep 7, 2023