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# In the figure, point D divides side BC of triangle ABC into

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Intern
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Re: In the figure, point D divides side BC of triangle ABC into [#permalink]

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12 Aug 2016, 20:28
1
This post was
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A GMAT way to answer the problem, the best way I can explain it. (click on the image and then in the "magnifying glass" at the top left to enlarge to full size)
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Re: In the figure, point D divides side BC of triangle ABC into [#permalink]

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23 Aug 2016, 21:54
vigrah wrote:
say angle CAB=y
since sum of angles in a triangleis 180\
x+y+45=180
x+y=135 equation 1

line AD is dividing BC in 2:1 ratio
hence
X+2/3Y+60=180
X+2/3y=120 equation 2
solving equation 1 &2 we get x=75

This DOES NOT give us the correct solution. There is a calculation mistake here. Using the calculation mentioned, we get x=90 which is incorrect. This basically implies that the statement
"line AD is dividing BC in 2:1 ratio - > therefore corresponding angles also get divided in the same proportion" is invalid. The only way to go forward is to follow the approach that Bunuel mentioned.

Hope this helps clear the confusion around this.

Thanks.
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Re: In the figure, point D divides side BC of triangle ABC into [#permalink]

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12 Nov 2016, 19:57
enigma123 wrote:

In the figure, point D divides side BC of triangle ABC into segments BD and DC of lengths 1 and 2 units respectively. Given that ÐADC = 60º and ÐABD = 45º, what is the measure of angle x in degrees? (Note:
Figure is not drawn to scale.)

A. 55
B. 60
C. 70
D. 75
E. 90

Any idea guys what will be the correct answer please? Also any idea how can I cut and paste the pictures in my post? Is it possible?

[Reveal] Spoiler:
Attachment:
The attachment Triangle.png is no longer available

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File comment: www.GMATinsight.com

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Re: In the figure, point D divides side BC of triangle ABC into [#permalink]

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15 Nov 2016, 21:36
Hi Bunuel
please clarify how we decide that triangle CAO is isoseles and AO and line AO is s.root of 3
Also how we decide that line OB is s.root of 3
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In the figure, point D divides side BC of triangle ABC into [#permalink]

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14 Feb 2017, 03:49
Bunuel wrote:
enigma123 wrote:

In the figure, point D divides side BC of triangle ABC into segments BD and DC of lengths 1 and 2 units respectively. Given that ÐADC = 60º and ÐABD = 45º, what is the measure of angle x in degrees? (Note:
Figure is not drawn to scale.)

(A) 55
(B) 60
(C) 70
(D) 75
(E) 90

Any idea guys what will be the correct answer please? Also any idea how can I cut and paste the pictures in my post? Is it possible?

Complete solution for all the angles is in the image below:

x=45+30=75.

Notes:
Sides with one blue segment crossing them are equal and sides with two blues segments crossing them are equal too.

CO is perpendicular to AD --> OD=1 (from 30-60-90 right triangle property as sides are in ratio $$1:\sqrt{3}:2$$) --> as OD=BD=1 then ODB is an isosceles triangle.

<CDO and <BDO are supplementary to each other (supplementary angles are two angles that add up to 180°), so <BDO=120 --> <DAB=180-(120+45)=15.

As ODB is an isosceles triangle --> <DOB=<DBO=30. <OBA=45-30=15 --> AOB is an isosceles triangle, so OA=OB. Also COB is an isosceles triangle, so CO=OB --> OA=CO=OB. So, AOC is an isosceles triangle --> <CAO=<OCA=45 (as <COA=90) --> x=45+30=75.

P.S. You can attach image files directly to the post.

[Reveal] Spoiler:
Attachment:
Triangle complete.PNG

Hello sir,
Can you explain me by what properties you arrived at the solution?
By which property you drew CO such that it meets at the point where BO also meets to form such isosceles triangles?

Last edited by ravi19012015 on 14 Feb 2017, 10:32, edited 1 time in total.
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In the figure, point D divides side BC of triangle ABC into [#permalink]

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14 Feb 2017, 10:04
Brunnel :

Could we have used a 45:45:90 for the triangle DOB by drawing DO perpendicular to AB. However, Doing this wont take us anywhere near the triangle ADC correct?

So. I guess my question is, Though I can understand by seeing values of sides and angles in triangles that I somehow have to use sides and angles combined to get my answer, and one of the most popular way of doing those in GMAT is using 45:45:90 or 30:60:90 triangles, but how do choose at one go which triangle should I use? What should be the thinking behind it? Which triangle to draw the perpendicular from among the trianlges ACD and ADB?
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Re: In the figure, point D divides side BC of triangle ABC into [#permalink]

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26 Jun 2017, 15:45
ODB is an isosceles triangle --> <DOB=<DBO=30. <OBA=45-30=15 --> AOB is an isosceles triangle, so OA=OB. Also COB is an isosceles triangle, so CO=OB --> OA=CO=OB. So, AOC is an isosceles triangle --> <CAO=<OCA=45 (as <COA=90) --> x=45+30=75.
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Re: In the figure, point D divides side BC of triangle ABC into [#permalink]

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09 Jul 2017, 16:04
navigator123 wrote:
enigma123 wrote:
In the figure, point D divides side BC of triangle ABC into segments BD and DC of lengths 1 and 2 units respectively. Given that ÐADC = 60º and ÐABD = 45º, what is the measure of angle x in degrees? (Note:
Figure is not drawn to scale.)

A. 55
B. 60
C. 70
D. 75
E. 90

Any idea guys what will be the correct answer please? Also any idea how can I cut and paste the pictures in my post? Is it possible?

Is it correct to tell that <CAD : <BAD = 2:1 because,
CD:DB = 2:1?

Even I have imagined in the same way and got 90 as the answer. I understood the explanation from the figures but my doubt is why can't <CAD = 30? when <DAB=15? If CD:DB=2:1, is it wrong to assume <CAD:<DAB is also 2:1??? please answer.

Thanks,
Uma
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Re: In the figure, point D divides side BC of triangle ABC into [#permalink]

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10 Jul 2017, 23:51
What is the source of this question?
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Re: In the figure, point D divides side BC of triangle ABC into [#permalink]

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10 Jul 2017, 23:57
rekhabishop wrote:
What is the source of this question?

The tag indicated that the source is Manhattan GMAT
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Re: In the figure, point D divides side BC of triangle ABC into [#permalink]

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20 Jul 2017, 23:54
Although I got the right answer, I wanted to understand simpler equation to derive the answer. Can someone help!
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Re: In the figure, point D divides side BC of triangle ABC into [#permalink]

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17 Aug 2017, 08:30
2
KUDOS
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. see attached pic

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Re: In the figure, point D divides side BC of triangle ABC into [#permalink]

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06 Dec 2017, 19:10
Hi bunuel,

Sorry for bumping. A quick question.

How do we know that AD is not equal to 1. If it is equal to 1 then the ratio 2:1 when a perpendicular is drawn does not hold good right? Am i missing something?
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Re: In the figure, point D divides side BC of triangle ABC into [#permalink]

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06 Dec 2017, 19:40
gauthamvm wrote:
Hi bunuel,

Sorry for bumping. A quick question.

How do we know that AD is not equal to 1. If it is equal to 1 then the ratio 2:1 when a perpendicular is drawn does not hold good right? Am i missing something?

x=45+30=75.

Notes:
Sides with one blue segment crossing them are equal and sides with two blues segments crossing them are equal too.

CO is perpendicular to AD --> OD=1 (from 30-60-90 right triangle property as sides are in ratio $$1:\sqrt{3}:2$$) --> as OD=BD=1 then ODB is an isosceles triangle.
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Re: In the figure, point D divides side BC of triangle ABC into [#permalink]

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07 Dec 2017, 08:43
Hi Bunuel
is this a GMAT level question? seems too tough
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Re: In the figure, point D divides side BC of triangle ABC into   [#permalink] 07 Dec 2017, 08:43

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