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Triangles ABC, ACD, and ABD are all isosceles triangles. Poi

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Triangles ABC, ACD, and ABD are all isosceles triangles. Poi [#permalink] New post 07 May 2020, 10:34
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70% (01:47) correct 29% (01:59) wrong based on 17 sessions
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Triangles ABC, ACD, and ABD are all isosceles triangles. Point E (not shown) is the midpoint of \(\overline{BD}\). If the ratio of the length of \(\overline{CE}\) to the length of \(\overline{BC}\) is equal to \(\sqrt{3} : 2\), then what is the measure, in degrees, of ∠CAD?

A. 10

B. 15

C. 30

D. 45

E. 60

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[Reveal] Spoiler: OA

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Re: Triangles ABC, ACD, and ABD are all isosceles triangles. Poi [#permalink] New post 07 May 2020, 20:11
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△ABC, △ACD and △ABD are isosceles;
In △ABC: AC=BC, ∠BAC = ∠ABC = x
△ACD: AC=CD, ∠CAD = ∠CDA = x
△ABD: AB=AD,

Now, In △BCE, CE/BC = √3/2
sin ∠CBE = length of oppo site/ length of hypotenus
CE / BC = √3/2 = Sin 60°
hence, ∠CBE = 60°
as BC = CD; ∠CDE = 60°


In △ABD; ∠A + ∠B + ∠C = 180
2x + (x+60) + (x+60) = 180
4x = 60
x = 15
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Re: Triangles ABC, ACD, and ABD are all isosceles triangles. Poi [#permalink] New post 12 Oct 2020, 07:47
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The correct answer is b) 15 degrees.

Now, let us work our way from the ratio CE/BC = √3:2 = > ECB is a 30-60-90 triangle.
=> Angle CBD = 60;
Angle CDB = 60
Now, notice that since two angles are 60 degree in Triangle CDB, CDB is equilateral
=> Angle DCB = 60 degree

Now notice that CE bisects the angle DCB

Hence, ACB + BCE = 180
=> ACB + 30 = 180
=> ACB = 150

Similarly,
ACD + DCE = 180
=> ACD + 30 = 180
=> ACD = 150

Now notice, that equal sides of the isosceles triangle DAC must be the sides AC and CD that are adjacent to angle ACD
(This is because if side opposite to ACD is the equal side, then 150 + 150 > 180 degrees. Hence not possible)

This ,means that CAD = CDA = x
=> x + x = 150 = 180
=> 2x = 30
=> x = 15
=> CAD = 15
Re: Triangles ABC, ACD, and ABD are all isosceles triangles. Poi   [#permalink] 12 Oct 2020, 07:47
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Triangles ABC, ACD, and ABD are all isosceles triangles. Poi

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