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Intern  Joined: 03 Jan 2016
Posts: 3
Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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here are my way to solve it:
Attachments mi-proceso.png [ 40.48 KiB | Viewed 7476 times ]

Intern  Joined: 03 Jan 2016
Posts: 3
Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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1
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)
Don't forget +1KUDOS !!!
Attachments triangle-problem.png [ 85.25 KiB | Viewed 7010 times ]

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

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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 segments  [#permalink]

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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?

Attachment:
The attachment Triangle.png is no longer available

Attachments

File comment: www.GMATinsight.com 12.jpg [ 110.35 KiB | Viewed 2484 times ]

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

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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?
Math Expert V
Joined: 02 Sep 2009
Posts: 59147
Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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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 segments  [#permalink]

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vigrah wrote:
say angle CAB=y
since sum of angles in a triangle is 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

Bunnel/Experts,

I have a doubt here.I understand angles and sides ratio will be proportional.

So If I see angle opposite to side with lenght 1 is 15 degree .So why not we
directly comprehend angle opposite to CB(3 units) is 45 degrees.

from this I will get angle y as 45 degrees
Math Expert V
Joined: 02 Sep 2009
Posts: 59147
Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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prabsahi wrote:
vigrah wrote:
say angle CAB=y
since sum of angles in a triangle is 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

Bunnel/Experts,

I have a doubt here.I understand angles and sides ratio will be proportional.

So If I see angle opposite to side with lenght 1 is 15 degree .So why not we
directly comprehend angle opposite to CB(3 units) is 45 degrees.

from this I will get angle y as 45 degrees

The solution you are quoting is not correct. Please check the discussion on previous pages.
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In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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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.

Attachment:
Triangle complete.PNG

Hi Bunuel,

I have two questions which came in when I was solving the problem. Tried reading through the other explanations but am still unclear.

1. When you drew the segment OB, how did you determine that it will split angle B (i.e. 45) into 30 and 15? Why could it not be any other split like 20 and 25 or 10 and 35 etc?
2. In triangle AOC we know that angle O is 90, but then how did you conclude that the other 2 angles are 45 each and not any other split?

Warm Regards,
Pritishd
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Cheers. Wishing Luck to Every GMAT Aspirant! In the figure, point D divides side BC of triangle ABC into segments   [#permalink] 13 Oct 2019, 09:33

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