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According to the figure above, does AD = BD?

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Bunuel
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According to the figure above, does AD = BD? [#permalink] New post   16 Jul 2015, 01:24
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A
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C
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E

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45% (medium)

Question Stats:

62% (01:44) correct 38% (02:21) wrong based on 175 sessions
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According to the figure above, does AD = BD?

(1) The degree measure of angle ABD is half that of angle BDC.
(2) The degree measure of angle DBC is 20°.


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A
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Re: According to the figure above, does AD = BD? [#permalink] New post   16 Jul 2015, 11:08
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Bunuel wrote:

According to the figure above, does AD = BD?

(1) The degree measure of angle ABD is half that of angle BDC.
(2) The degree measure of angle DBC is 20°.


Kudos for a correct solution.

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The attachment triangles.gif is no longer available


Check the working in the figure attached

Answer: Option A
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Re: According to the figure above, does AD = BD? [#permalink] New post   16 Jul 2015, 14:55
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We want to know if AD = BD.
If true <ABD would = <DAB

Stmt 1 tells us that <ABD is 1/2 of <BDC.
- Let's call <ABD x and <BDC 2x.
- If <BDC is 2x, then <BDA is 180-2x
- <DAB is then 180 - x - (180-2x) ==> or, <DAB = x
- Since <ABD and <DAB both equal x, Stmt 1 is SUFFICIENT
- Answer Choice is either A or D

Stmt 2 tells us that <DBC is 20
- We cannot determine anything further from this information
- Not sufficient

Answer is A
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According to the figure above, does AD = BD? [#permalink] New post   16 Jul 2015, 03:31
Bunuel wrote:

According to the figure above, does AD = BD?

(1) The degree measure of angle ABD is half that of angle BDC.
(2) The degree measure of angle DBC is 20°.


Kudos for a correct solution.

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Attachment:
triangles.gif


Statement 2 is obviously not sufficient.

Per statement 1, in Triangle ABD, let \(\angle{ABD} = X ----> \angle{BDC} = 2X.\)
By external angle theorem, \(\angle{BAD} + \angle{ABD} = \angle{BDC} ----> \angle{BAD} = \angle{BDC} - \angle{ABD} = 2X-X = X ----> \angle{BAD} = \angle{ABD}\)

Thus AD=BD ---> This statement is sufficient. A is the correct answer.
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Re: According to the figure above, does AD = BD? [#permalink] New post   17 Jul 2015, 15:44
Bunuel wrote:

According to the figure above, does AD = BD?

(1) The degree measure of angle ABD is half that of angle BDC.
(2) The degree measure of angle DBC is 20°.


Kudos for a correct solution.

Show Spoiler
Attachment:
triangles.gif



St 1:
Given \(\angle ABD\) = 1/2 * \(\angle BDC\)
Assume \(\angle ABD\) = x then, \(\angle BDC\) = 2x

Sum of the Angles in Triangle ABD => \(\angle BAD\) + x + 180 - 2x = 180;
Solving \(\angle BAD\) = x
=> \(\angle BAD\) =\(\angle ABD\). Hence ABD is an isoceles traingle with sides AD = BD.

Hence Sufficient.

St 2: Given angle DBC is 20°. Cannot infer anything regarding to sides AD = BD. Hence Not Sufficient.

Option A.
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Re: According to the figure above, does AD = BD? [#permalink] New post   19 Jul 2015, 13:49
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Bunuel wrote:

According to the figure above, does AD = BD?

(1) The degree measure of angle ABD is half that of angle BDC.
(2) The degree measure of angle DBC is 20°.


Kudos for a correct solution.

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Attachment:
triangles.gif


800score Official Solution:

Using Statement (1), let's call angle BDC = x. Now, angle BDA = 180° – x and angle ABD = x/2.
Furthermore, all three angles in triangle ABD must sum to 180°, so we can write the following equation:
180° = x/2 + (180° – x) + angle BAD.
Solving this equation, we find that angle BAD = x/2, which is the same degree measure as that of angle ABD. Since these two angles are the same, the triangle is isosceles, and AD = BD. Statement (1) is sufficient.

Using Statement (2) alone, we cannot determine any relationships between the other angles, so Statement (2) is insufficient.

Since Statement (1) is sufficient and Statement (2) is not, the correct answer is choice (A).
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Re: According to the figure above, does AD = BD? [#permalink] New post   27 Sep 2016, 07:45
Bunuel wrote:

According to the figure above, does AD = BD?

(1) The degree measure of angle ABD is half that of angle BDC.
(2) The degree measure of angle DBC is 20°.


Kudos for a correct solution.

Show Spoiler
Attachment:
triangles.gif


damn..it's been a while since I last time solved geometry questions...
took way too long to re-check everything...

1. let's say ABD is X. BDC is 2x. Since we know that angle BDC is 2x (two times the angle ABD), we know for sure that angle ADB is 180-2x (property of the lines).
let's name BAD angle Y.
we have: 180 = x+180-2x+y
rearrange = x=y.
it means that angle ABD is equal to BAD. As such, AD is equal to BD. sufficient.

2. doesn't give much of the info alone...

A is the answer.
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Re: According to the figure above, does AD = BD? [#permalink] New post   27 Sep 2016, 10:35
Statement 1 sufficient ABD = x , BDC = 2x. We can conclude ADB = 180-2x and BAD = 180-(180-2x)-x = x so BAD=ABD therefore the two opposite side are equal
Statement 2 insufficient DBC = 20 BDC and BCD could have multiple values
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Re: According to the figure above, does AD = BD? [#permalink] New post   20 Mar 2023, 08:38
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Re: According to the figure above, does AD = BD? [#permalink]
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