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

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

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04 Feb 2015, 13:54
So easy if you look at it the right way...
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Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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06 Mar 2015, 17:47
enigma123 wrote:
Attachment:
Triangle.png

Guys,
Is there any relation between the ratio of the sides to the angle containing them ?

i. e from the diagram ,$$\frac{CD}{DB}$$ = ratio of $$\angle CAD$$ and $$\angle DAB$$ ?
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Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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10 May 2015, 12:24
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

According to your calculation x=90 not 75
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Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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16 May 2015, 06:44
@Bunnel

Can you please explain how this is derived?

"As 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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18 May 2015, 10:45
1
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
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Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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18 May 2015, 10:50
3
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 4168 times ]

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

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28 May 2015, 11:37
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

That looks like a very efficient way to solve this. Could you explain the red formatted part? How do you set up this equation?
Thank you
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Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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28 May 2015, 11:39
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
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Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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28 May 2015, 20:13
1
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
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Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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28 May 2015, 20:34
<DAB can be calculated to be 15 degree
The side AD divides <CAB in the same ratio as it divides the side CB i.e. 2:1
Thus - <CAD = 30 degree and x = 90 degree
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Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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12 Jul 2015, 06:11
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?

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

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12 Jul 2015, 06:55
1
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?

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

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27 Oct 2015, 18:39
bunnuel's explanation is brilliant.
I did a mistake, and drew a perpendicular to CD, and concluded that angle A must be 60...
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Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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24 Nov 2015, 20:04
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

x+y = 135---1
x+2/3y = 120----2

subtracting 2 from 1

1/3 y = 15

y =45.

Substituting y =45 in eq 1
x+45 = 135
x = 90.

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

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07 Mar 2016, 03:35
Let me provide you an easy approach for this question:

Using side & angle ratio to determine x

assume angle DAC = y

then, angle (BAD/BAC) = side (BD/BC)

=> 15/y = 1/(1+2)
=> 15/y = 1/3
=> 15*3 = y
=>45

Now, as we know sum of all angles of a triangle is 180
Therefore, in triangle ADC, Angle (DAC+ACD+CDA) = 180
=> 60+45+X = 180
=> X = 75
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Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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15 Mar 2016, 10:13
Since the AD divides the BC in ration 2:1 , it means if we extend the traingle and make it eqilateral triangle. The medians of equilateral triangles divides each other in ration 2:1 . Thus easily all sides are 60 degrees and the x = 75. ..Simple. Please give Kudos .
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Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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Updated on: 12 Aug 2016, 19:26
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.
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example2.png [ 13.01 KiB | Viewed 2794 times ]

Originally posted by OmnerLV on 11 Aug 2016, 13:11.
Last edited by OmnerLV on 12 Aug 2016, 19:26, edited 1 time in total.
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Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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11 Aug 2016, 15:19
here are my way to solve it:
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mi-proceso.png [ 40.48 KiB | Viewed 2727 times ]

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

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12 Aug 2016, 20:28
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 !!!
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triangle-problem.png [ 85.25 KiB | Viewed 2238 times ]

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Re: In the figure, point D divides side BC of triangle ABC into segments  [#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 segments &nbs [#permalink] 23 Aug 2016, 21:54

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