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# Triangle

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In the diagram below, we have \$\angle ABC = \angle CAB = \angle DEB=\angle BDE.\$ Find \$AB\$ and \$CD\$.

Dec 20, 2023

#1
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Well, DEB is isoceles, so EB=3. then, i don't know what to do, but i have a feeling that be and ad are angle bisectors, but i can't prove it???

Dec 20, 2023
#2
+129690
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Since  angle BDE = angle DEB  then EB = DB = 3

Using the Law of Cosines

EB^2 = DB^2 + ED^2 - 2(DB)*ED) cos BDE

3^2 = 3^2 + 2^2 - 2(3)(2) cos BDE

[ 3^2 - 3^2 -2^2] / [ -2 * 3 * 2 ]  =  cos BDE

1/3 = cos BDE

And angle ABD = angle BDE =angle BDA

Let AE = x

AB = 2 + x

By the Law of Cosines again

AD^2  = AB^2  + DB^2 - 2(AB)(DB)cos DBA

(2 +x)^2  = (2 + x)^2 + 3^2 - 2(2+x)(3)(1/3)

-3^2 = -2(2 + x)

-9  = -4 - 2x

-5 = -2x

x = 5/2  = AE

So AB = 2 + x =  2 + 5/2 =  9/2 =  4.5

Let CD = y

So BC = 3 + y = AC    (remember that angle ABC  =  angle CAB)

Law of Cosines again

BC^2  = AB^2 + AC^2 - 2 ( AB * AC) cos CAB        (cos CAB  = cos BDE)

(3 + y)^2  = 4.5^2 + (3 + y)^2 - 2 [ (3+y) (4.5) ] (1/3)

4.5^2 = 3 (3 + y)

20.25 = 9 + 3y

11.25 / 3  = y   =  3.75  =  CD

Dec 20, 2023