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In the figure shown, AC = 2 and BD = DC = 1. What is the measure of  [#permalink]

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Attachment: 2018.OG.05.026.q.png [ 32.11 KiB | Viewed 17607 times ]

In the figure shown, AC = 2 and BD = DC = 1. What is the measure of angle ABD?

A. 15°
B. 20°
C. 30°
D. 40°
E. 45°

Originally posted by AbdurRakib on 17 Jun 2017, 10:21.
Last edited by Bunuel on 17 Jun 2017, 13:07, edited 1 time in total.
Renamed the topic and edited the question.
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In the figure shown, AC = 2 and BD = DC = 1. What is the measure of  [#permalink]

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AbdurRakib wrote:
Attachment:
2018.OG.05.026.q.png

In the figure shown, AC = 2 and BD = DC = 1. What is the measure of angle ABD?

A. 15°
B. 20°
C. 30°
D. 40°
E. 45°

If AC = 2 and DC = 1, then we can conclude that AD = 1
Add this information to the diagram... If AD = 1 and BD = 1, then ∆ABD is an ISOSCELES triangle, which means ∠DAB = ∠ABD are EQUAL... If we let x = BOTH ∠DAB and ∠ABD, then we can use the fact that angles in a triangle add to 180º
We can write: x + x + 120 = 180
Simplify: 2x + 120 = 180
Solve: x = 30

In other words, ∠ABD = 30º

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Originally posted by BrentGMATPrepNow on 03 Jul 2017, 11:14.
Last edited by BrentGMATPrepNow on 16 Apr 2018, 11:40, edited 1 time in total.
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Re: In the figure shown, AC = 2 and BD = DC = 1. What is the measure of  [#permalink]

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AbdurRakib wrote:
Attachment:
2018.OG.05.026.q.png
A. 15°
B. 20°
C. 30°
D. 40°
E. 45°

Given BD = DC = 1

AC = 2.

$$\triangle$$ ABD and $$\triangle$$ BDC are isosceles triangles.

$$\triangle$$ ABD is isosceles, sides AD = BD = 1. Therefore $$\angle$$ BAD = $$\angle$$ ABD. Let this angle be x.

Given $$\angle$$ ADB is $$120^{\circ}$$

$$\angle$$ ADB + $$\angle$$ BAD + $$\angle$$ ABD = 180

$$120 + x + x = 180$$

$$2x = 180 - 120 = 60$$

$$x = \frac{60}{2} = 30^{\circ}$$. Answer (C)...
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Re: In the figure shown, AC = 2 and BD = DC = 1. What is the measure of  [#permalink]

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Re: In the figure shown, AC = 2 and BD = DC = 1. What is the measure of  [#permalink]

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Angle BDC is 60 degree(because the angle in a straight line is equal to 180 degree)
Since BD = DC(as given in the question stem) the triangle must be equilateral.
Therefere DC=1.

Since AC=2, AD = AC - DC = 1
Angle ABD = Angle DAB = x(Angles opposite equal sides are equal)

We know that the sum of angles in a triangle is 180 degree,
2x + 120 = 180
Angle ABD(x) = 30 degree(Option C)
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Re: In the figure shown, AC = 2 and BD = DC = 1. What is the measure of  [#permalink]

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Sum of interior opposite angles = exterior angle.
$$\angle BAD + \angle ABD = \angle BDC$$
Here these two angles are equal, lets say x.

Since AC is a straight line, $$\angle BDC = 180^ {\circ} - 120^ {\circ} = 60^ {\circ}$$

Hence $$\angle BAD + \angle ABD = 60^ {\circ}$$
x+x=60
x=30

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Re: In the figure shown, AC = 2 and BD = DC = 1. What is the measure of  [#permalink]

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AbdurRakib wrote:
Attachment:
2018.OG.05.026.q.png

In the figure shown, AC = 2 and BD = DC = 1. What is the measure of angle ABD?

A. 15°
B. 20°
C. 30°
D. 40°
E. 45°

Since AC = 2 and DC = 1, AD must be 1. Since BD = 1, this makes triangle ABD an isosceles triangle with angle D as the vertex angle and angles A and ABD as the base angles. We know that angles A and ABD are equal because the sides opposite them are equal. If we let each of the base angles = x, we can create the following equation:

x + x + 120 = 180

2x = 60

x = 30

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In the figure shown, AC = 2 and BD = DC = 1. What is the measure of  [#permalink]

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Can someone help me, I have a flaw in my logic.
I totally understand it, but I would have solved it from the second triangle.

Is it true that angles in the triangle on the right are 60 degree?

I somehow got stuck because I thought that if those angles are 60, the upper angle (aka the one we are looking for) has to be 120.
Is this wrong because it would only be 120 if I would draw a line from up there and the touch on the line would make it 180?

See drawing:
Attachments aaa.PNG [ 249.06 KiB | Viewed 4758 times ]

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Re: In the figure shown, AC = 2 and BD = DC = 1. What is the measure of  [#permalink]

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chrtpmdr wrote:
Can someone help me, I have a flaw in my logic.
I totally understand it, but I would have solved it from the second triangle.

Is it true that angles in the triangle on the right are 60 degree?

I somehow got stuck because I thought that if those angles are 60, the upper angle (aka the one we are looking for) has to be 120.
Is this wrong because it would only be 120 if I would draw a line from up there and the touch on the line would make it 180?

See drawing:

Are you saying angle ABC is 120 as well? That's where the error is. You can solve this many ways: the easy way is to realize that angle BDC is 60(linear pair). AC= 2 (given).If DC=1, then AD= 1 (AD+DB=AC). That means triangle ADB is an isosceles triangle with base angles DAB and DBA congruent. angle DAB + angle DBA+120=180. From there you get angle DBA is equal to 30 degrees.

Now if you wanted to use the triangle on the right, you have correctly deduced that the triangle on the right is equiangular. Note that segment BD is the median of triangle ABC. Since AD=BD=DC, we know that the triangle ABC is a right triangle. (the median on the hypotenuse of a right triangle divides the triangle into two isosceles triangles. That means ABC is a right angle. Since DBC is 60 degrees, then ABD is 30 degrees. Re: In the figure shown, AC = 2 and BD = DC = 1. What is the measure of   [#permalink] 18 Jun 2020, 23:24

# In the figure shown, AC = 2 and BD = DC = 1. What is the measure of  