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In the diagram above, ∠A = ∠ABC, ∠CBD = ∠BDC, and ∠CBE =90°. If AE =
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26 Feb 2015, 06:48
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Re: In the diagram above, ∠A = ∠ABC, ∠CBD = ∠BDC, and ∠CBE =90°. If AE =
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26 Feb 2015, 11:30
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 1802Y. Angle C supplementary ,1802x+1802y=180 2x+2y=180 x+y =90 In trianlge BDE ANGLE B 90y. angle D 180y anlge E 2Y90= 2yxy =yx. 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.



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In the diagram above, ∠A = ∠ABC, ∠CBD = ∠BDC, and ∠CBE =90°. If AE =
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26 Feb 2015, 18:59
Answer = B = 8 Refer diagram below Attachment:
differenceofsquares.png [ 25.12 KiB  Viewed 2390 times ]
∠A = ∠ABC = ∠CBD = ∠BDC (Angles are same, so corresponding lengths are same & viceversa) AC = CB = CD = 6 BE \(= \sqrt{(6+4)^2  6^2} = \sqrt{64} = 8\)
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Re: In the diagram above, ∠A = ∠ABC, ∠CBD = ∠BDC, and ∠CBE =90°. If AE =
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02 Mar 2015, 05:35
Bunuel wrote: Attachment: differenceofsquares.png In the diagram above, ∠A = ∠ABC, ∠CBD = ∠BDC, and ∠CBE =90°. If AE = 16 and DE = 4, what is the length of BE? (A) 7 (B) 8 (C) 9 (D) 10 (E) 11 Kudos for a correct solution. 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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Re: In the diagram above, ∠A = ∠ABC, ∠CBD = ∠BDC, and ∠CBE =90°. If AE =
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18 Jun 2017, 07:05
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 306090, 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



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Re: In the diagram above, ∠A = ∠ABC, ∠CBD = ∠BDC, and ∠CBE =90°. If AE =
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02 Mar 2019, 13:05
PareshGmat wrote: Answer = B = 8 Refer diagram below Attachment: differenceofsquares.png ∠A = ∠ABC = ∠CBD = ∠BDC (Angles are same, so corresponding lengths are same & viceversa) AC = CB = CD = 6 BE \(= \sqrt{(6+4)^2  6^2} = \sqrt{64} = 8\) Isn't it wrong to assume that ∠A = ∠ABC = ∠CBD = ∠BDC We know their corresponding sides are equal, but that does not necessarily mean they are equal. Also, since it is drawn to scale, it is evident they are not equal, since if they were, the triangles would be congruent based on AAS rule. Another confirmation they are not equal is because if they were, then they would each have to be equal to 45oC,which means ∠ACB = 90 and ∠BCD =90 which is evidently wrong




Re: In the diagram above, ∠A = ∠ABC, ∠CBD = ∠BDC, and ∠CBE =90°. If AE =
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02 Mar 2019, 13:05






