In the diagram above, ∠A = ∠ABC, ∠CBD = ∠BDC, and ∠CBE =90°. If AE =

Answer = B = 8

Refer diagram below



∠A = ∠ABC = ∠CBD = ∠BDC (Angles are same, so corresponding lengths are same & vice-versa)

AC = CB = CD = 6

BE \(= \sqrt{(6+4)^2 - 6^2} = \sqrt{64} = 8\)

MAGOOSH OFFICIAL SOLUTION:

This is a tricky one. Remember, it’s a diagram drawn to scale, so if all else fails, you can estimate (see this post). But, let’s solve this with math. The fact that ∠A = ∠ABC tells us triangle ABC is isosceles, with AC = BC. The fact that ∠CBD = ∠BDC tells us triangle BCD is isosceles, with BC = CD. The fact that ∠CBE = 90° means that (BC)2 + (BE)2 = (CE)2. This means

(BE)^2 = (CE)^2 - (BC)^2
= (CE + BC)(CE - BC)
= (CE + AC)(CE - CD) (substitutions from the two isosceles triangles)
= (AE)(DE)
= (16)(4) = 64 --> BE = 8

Answer = B.
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in triangle abc Angle A= Angle ABC. Both sides are equal. put angle a = X
ANGLE C 180- 2X
In triangle BDC. ANGLE B= ANGLE D.put angle B = Y
anlge C 180-2Y.
Angle C supplementary ,180-2x+180-2y=180
2x+2y=180
x+y =90
In trianlge BDE ANGLE B 90-y.
angle D 180-y
anlge E 2Y-90= 2y-x-y
=y-x. so y >=x
if y = 45 degree , then in Tri BDE , ang B =45, ang D 135 can"t satisfy
If y = 50 degree , ang B 40, ang D 130, ange E 10. hen In Triangle ABC ang A 40, angle B 130, angle C 10

If y = 55 degree, ang B 35, ang D 125 , ang E35 cant satisfy.

if y = 60 degre, ang B 30, ang D 120, ang E 30 satisfy. only y = 50 ,60, 70 ,80 satisfy. then Here also Angle A 30, angle B 120, Angle C = 30.
Triangle ABC and Triangle BDE are congruent.


BE/AE = DE/BE => BE^2=AE*DE= 16*4=64.
BE = 8.

Answer is B.

I've got another way to solve the problem.

You know line segment DE is 4.
First step: Form a triangle with a line with an angle of 30 degrees, from point D, to line segment BE. Let's call this new point, point F.
Point F now has 2 right angles.
Second step: Triangle DEF is a triangle with angles of 30-60-90, so proportions are x : xscrt3 : 2x
Fill in (since 2x is 4), 2 : 2sqrt3 : 4
(In triangle DEF, line segment DF is the long leg)
Now you now that line segment DF is 2sqrt3.
Third step: Triangle BDF also has a right triangle and you know the short leg is 2sqrt3. (notice how the long leg from DEF is the short leg in BDF)
Same proportions x : xsqrt3 : 2x
Fill in to find the long leg: (2sqrt3)*sqrt3=6
Last step: Add up line segments BF and EF --> 6+2= 8

This was rather easy.

∠A = ∠ABC (given) ----> AC = BC
∠CBD = ∠BDC, ----> BC = CD

Therefore, AC = BC = CD = 6

CE = DC + DE
CE = 6 + 4 = 10

CE is the hypotenuse of rt angled tr. BCE

BC^2 + BE ^2 = CE ^2
6^2 + BE ^2 = 10^2

Therefore, BE = 8

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