Last visit was: 24 Jul 2024, 10:54 It is currently 24 Jul 2024, 10:54
Toolkit
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

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

SORT BY:
Tags:
Show Tags
Hide Tags
Senior Manager
Joined: 25 Jun 2011
Status:Finally Done. Admitted in Kellogg for 2015 intake
Posts: 395
Own Kudos [?]: 17063 [669]
Given Kudos: 217
Location: United Kingdom
Concentration: International Business, Strategy
GMAT 1: 730 Q49 V45
GPA: 2.9
WE:Information Technology (Consulting)
Math Expert
Joined: 02 Sep 2009
Posts: 94606
Own Kudos [?]: 643590 [229]
Given Kudos: 86737
Manager
Joined: 10 Jul 2013
Posts: 228
Own Kudos [?]: 1050 [42]
Given Kudos: 102
General Discussion
Manager
Joined: 22 Dec 2011
Posts: 174
Own Kudos [?]: 1062 [0]
Given Kudos: 32
Re: In the figure, point D divides side BC of triangle ABC into segments [#permalink]
Bunuel wrote:
Complete solution for all the angles is in the image below:
Attachment:
Triangle complete.PNG
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.

Bunuel - Can we not use exterior angle property

for triangle ACD
Angle at C = x
<CDA = 60
Since sum of all the angle of triangle is 180 then we

<CAD + <CDA + x = 180
<CAD = 180 -60 -x
<CAD = 120 -x

For the other triangle we know
<DAC = 15
<DBA =45

Exterior angle is sum of opp interior angle and not adjacent to it
120 -x = 120 + 45
x = 45.

Since this is not the ans could you please let me know what I'm doing wrong here?

Many thanks.

Cheers
Math Expert
Joined: 02 Sep 2009
Posts: 94606
Own Kudos [?]: 643590 [1]
Given Kudos: 86737
Re: In the figure, point D divides side BC of triangle ABC into segments [#permalink]
1
Bookmarks
Jp27 wrote:
Bunuel wrote:
Complete solution for all the angles is in the image below:
Attachment:
The attachment Triangle complete.PNG is no longer available
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.

Bunuel - Can we not use exterior angle property

for triangle ACD
Angle at C = x
<CDA = 60
Since sum of all the angle of triangle is 180 then we

<CAD + <CDA + x = 180
<CAD = 180 -60 -x
<CAD = 120 -x

For the other triangle we know
<DAC = 15
<DBA =45

Exterior angle is sum of opp interior angle and not adjacent to it
120 -x = 120 + 45
x = 45.

Since this is not the ans could you please let me know what I'm doing wrong here?

Many thanks.

Cheers

<CAD is not exterior angle of <DAB. Exterior angle for <DAB is given below in green:
Attachment:

Exterior.png [ 44.25 KiB | Viewed 169323 times ]
Hope it's clear.
Intern
Joined: 15 May 2013
Posts: 5
Own Kudos [?]: 96 [87]
Given Kudos: 7
In the figure, point D divides side BC of triangle ABC into segments [#permalink]
66
Kudos
21
Bookmarks
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

Edit: THIS SOLUTION IS NOT CORRECT. IGNORE IT
Manager
Joined: 24 Nov 2012
Posts: 131
Own Kudos [?]: 1022 [0]
Given Kudos: 73
Concentration: Sustainability, Entrepreneurship
GMAT 1: 770 Q50 V44
WE:Business Development (Internet and New Media)
Re: In the figure, point D divides side BC of triangle ABC into segments [#permalink]
Bunuel wrote:
--> 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.

.

I lost you from here. How do you know that angle cab is 45+15?
Math Expert
Joined: 02 Sep 2009
Posts: 94606
Own Kudos [?]: 643590 [1]
Given Kudos: 86737
Re: In the figure, point D divides side BC of triangle ABC into segments [#permalink]
1
Kudos
Transcendentalist wrote:
Bunuel wrote:
--> 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.

.

I lost you from here. How do you know that angle cab is 45+15?

<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.

<CAB = <DAB + <CAO = 15 + 45.
Manager
Joined: 04 Jan 2014
Posts: 84
Own Kudos [?]: 56 [2]
Given Kudos: 24
Re: In the figure, point D divides side BC of triangle ABC into segments [#permalink]
1
Kudos
hi Bunnel,

How can we be sure that AO is equals to CO?
Math Expert
Joined: 02 Sep 2009
Posts: 94606
Own Kudos [?]: 643590 [2]
Given Kudos: 86737
Re: In the figure, point D divides side BC of triangle ABC into segments [#permalink]
1
Kudos
1
Bookmarks
pretzel wrote:
hi Bunnel,

How can we be sure that AO is equals to CO?

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.
Intern
Joined: 11 Sep 2010
Status:Active
Posts: 4
Own Kudos [?]: 5 [0]
Given Kudos: 96
Location: United States
Concentration: General Management, Technology
WE:Information Technology (Health Care)
Re: In the figure, point D divides side BC of triangle ABC into segments [#permalink]
@Bunnel

Can you please explain how this is derived?

"As ODB is an isosceles triangle"
Senior Manager
Joined: 07 Aug 2011
Posts: 422
Own Kudos [?]: 1799 [6]
Given Kudos: 75
Concentration: International Business, Technology
GMAT 1: 630 Q49 V27
Re: In the figure, point D divides side BC of triangle ABC into segments [#permalink]
5
Kudos
1
Bookmarks
ashtitude wrote:
@Bunnel

Can you please explain how this is derived?

"As ODB is an isosceles triangle"

May be the attached image can help explain how.
Attachments

IMG-20150514-WA0014.jpg [ 44.46 KiB | Viewed 70739 times ]

Retired Moderator
Joined: 29 Apr 2015
Posts: 717
Own Kudos [?]: 4237 [0]
Given Kudos: 302
Location: Switzerland
Concentration: Economics, Finance
Schools: LBS MIF '19
WE:Asset Management (Investment Banking)
Re: In the figure, point D divides side BC of triangle ABC into segments [#permalink]
EgmatQuantExpert wrote:
ashtitude wrote:
@Bunnel

Can you please explain how this is derived?

"As ODB is an isosceles triangle"

Hi ashtitude,

Using image drawn by Bunuel for explanation here.

You can observe in the figure that triangle OCD is a 30-60-90 triangle, hence it sides will be in the ratio 1:√3:2. Since CD = 2, we will have OD = 1. So in triangle OBD we have OD = BD = 1 which makes ODB an isosceles triangle.

Hope this helps

Regards
Harsh

Nice explanation. What confuses me more is, how do we know that AOB is isosceles?
Thank you
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3710
Own Kudos [?]: 17362 [2]
Given Kudos: 165
Re: In the figure, point D divides side BC of triangle ABC into segments [#permalink]
2
Kudos
reto wrote:
EgmatQuantExpert wrote:
ashtitude wrote:
@Bunnel

Can you please explain how this is derived?

"As ODB is an isosceles triangle"

Hi ashtitude,

Using image drawn by Bunuel for explanation here.

You can observe in the figure that triangle OCD is a 30-60-90 triangle, hence it sides will be in the ratio 1:√3:2. Since CD = 2, we will have OD = 1. So in triangle OBD we have OD = BD = 1 which makes ODB an isosceles triangle.

Hope this helps

Regards
Harsh

Nice explanation. What confuses me more is, how do we know that AOB is isosceles?
Thank you

Hi reto,

In triangle ADB, ∠ADB = 120 ( as∠ADB = 180 - ∠CDO = 180 - 60 =120). We now know two angles of the triangle ADB. So the third angle ∠DAB = 180 -120 -45 = 15.

Similarly in triangle ODB since ∠DBO =30 (as ODB is an isosceles triangle), we have ∠OBA = 45 - 30 = 15.

So we see that ∠OBA = ∠DAB = 15. These are two angles of triangle AOB. Hence AOB is an isosceles triangle with sides OA = OB

Hope it's clear

Regards
Harsh
Manager
Joined: 04 May 2015
Posts: 64
Own Kudos [?]: 30 [0]
Given Kudos: 58
Concentration: Strategy, Operations
WE:Operations (Military & Defense)
Re: In the figure, point D divides side BC of triangle ABC into segments [#permalink]
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

This looks much more simple!

Just so I am clear because line AD bisects line CB in a two to 1 ratio we can calculate the breakdown of angles at vertex A?

Thanks for your help
SVP
Joined: 20 Mar 2014
Posts: 2356
Own Kudos [?]: 3650 [1]
Given Kudos: 816
Concentration: Finance, Strategy
GMAT 1: 750 Q49 V44
GPA: 3.7
WE:Engineering (Aerospace and Defense)
Re: In the figure, point D divides side BC of triangle ABC into segments [#permalink]
1
Kudos
DropBear 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

This looks much more simple!

Just so I am clear because line AD bisects line CB in a two to 1 ratio we can calculate the breakdown of angles at vertex A?

Thanks for your help

This is a very dangerous inference. The only formula that links sides to angles in by sine law from trigonometry that says:

In a triangle ABC with sides a,b,c opposite $$\angle (A), \angle (B) and \angle (C)$$ respectively, then

a/sin(A) = b/sin(B) = c/sin(C). You do not have to remember this 'sine law' for GMAT.

It was a lucky coincidence that the user was able to arrive at the correct answer by assuming that the angles themselves got divided into 1:2 ratio if the sides were divided in the ratio 1:2.

The best way would be to follow what Bunuel did in his approach and for obtaining $$\angle (BAD) and \angle (CAD)$$, after you find OD = DB = 1 ---> $$\angle {DBO} = \angle {DOB}$$ = 30 degrees.

Also, as we know $$\angle{ABD}$$ = 45 degrees ---->$$\angle {ABO}$$ = 15 degrees. Additionally, $$\angle{ADB} = \angle {ADB} + \angle {DBO}$$ ... (by external angle theorem of triangles).

Once you do this, you will see $$\angle{BAO}$$ = 15 degrees. Then , you proceed to finding other angles as stated by Bunuel.
Intern
Joined: 03 Jan 2016
Posts: 3
Own Kudos [?]: 10 [1]
Given Kudos: 0
Re: In the figure, point D divides side BC of triangle ABC into segments [#permalink]
1
Kudos
We are solving it with the misconception that if we make a perpendicular from the point C to the line AD will perfectly connect with the point O(M). In this case fortunately matched, but if you try to use the same logic just modifying the value of line CD to 3, this method won't work. We need to find a real answer.

-UPDATE-

before start working with the original triangle, we need to understand the equilateral triangle properties. I'll explain later.
Attachments

example2.png [ 13.01 KiB | Viewed 68634 times ]

Originally posted by OmnerLV on 11 Aug 2016, 14:11.
Last edited by OmnerLV on 12 Aug 2016, 20:26, edited 1 time in total.
Intern
Joined: 03 Jan 2016
Posts: 3
Own Kudos [?]: 10 [1]
Given Kudos: 0
Re: In the figure, point D divides side BC of triangle ABC into segments [#permalink]
1
Kudos
here are my way to solve it:
Attachments

mi-proceso.png [ 40.48 KiB | Viewed 68545 times ]

Intern
Joined: 03 Jan 2016
Posts: 3
Own Kudos [?]: 10 [8]
Given Kudos: 0
Re: In the figure, point D divides side BC of triangle ABC into segments [#permalink]
5
Kudos
3
Bookmarks
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 68406 times ]

Manager
Joined: 19 Oct 2012
Posts: 221
Own Kudos [?]: 547 [0]
Given Kudos: 103
Location: India
Concentration: General Management, Operations
GMAT 1: 660 Q47 V35
GMAT 2: 710 Q50 V38
GPA: 3.81
WE:Information Technology (Computer Software)
Re: In the figure, point D divides side BC of triangle ABC into segments [#permalink]
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.
Re: In the figure, point D divides side BC of triangle ABC into segments [#permalink]
1   2
Moderator:
Math Expert
94606 posts