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Manager  Joined: 29 May 2008
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If A is the center of the circle shown above (see attachment  [#permalink]

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Difficulty:   45% (medium)

Question Stats: 69% (02:03) correct 31% (02:04) wrong based on 906 sessions

### HideShow timer Statistics Attachment: Circle - original.PNG [ 3.31 KiB | Viewed 26274 times ]
If A is the center of the circle shown above and AB = BC = CD, What is the value of x?

A. 15
B. 30
C. 45
D. 60
E. 75

Originally posted by TheRob on 07 Sep 2009, 07:19.
Last edited by Bunuel on 21 Sep 2013, 13:18, edited 2 times in total.
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Re: If A is the center of the circle shown above  [#permalink]

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TheRob wrote:
If A is the center of the circle shown above (see attachment) and AB = BC = CD, What is the value of x?

A. 15
B. 30
C. 45
D. 60
E. 75
Attachment:
The attachment Circle - original.PNG is no longer available

(I tried to draw it the best I could sorry for the inconvenience)

Look at the diagram below:
Attachment: Circle.PNG [ 26.39 KiB | Viewed 26532 times ]
Given that AB = BC = CD, also since AB is the radius then AB = AC = AD = radius, so we have that: AB = BC = CD = AC = AD, so basically we have two equilateral triangles ABC and ACD with common base of AC (ABC and ACD are mirror images of each other). Line segment BD cuts the angle ABC in half and since all angles in equilateral triangle equal to 60 degrees then x=60/2=30 degrees.

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Triangle abc is equilateral, therefore all angles are 60. abc = acb = bac = 60.
Similarly, Triangle adc is equilateral, therefore all angles are 60. adc = acd = dac = 60.

Now, angle bad = bac + dac = 120, since each triangle is a reverse image of the other.

Hence, B
##### General Discussion
Intern  Joined: 03 Sep 2009
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A is the center of the circle => AB= AD = AC = BC = CD
=> ABC is a regular triangle => x = 30°

TheRob wrote:
If A is the center of the circle shown above (see attachment) and AB = BC = CD, What is the value of x?
A) 15
B) 30
C) 45
D) 60
E) 75

(I tried to draw it the best I could sorry for the inconvenience)
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1
1
you can solve the question easily by knowing two important properties of square and rhombus.

1. diagonals bisect each other.
2. each diagonal forms a 90 degree angle with the other diagonal
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Re: If A is the center of the circle shown above  [#permalink]

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bunuel ,

how can we say that line segment BD cuts the angle ABC in half which property is this , can u expalin me?
thanks
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Re: If A is the center of the circle shown above  [#permalink]

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pbull78 wrote:
bunuel ,

how can we say that line segment BD cuts the angle ABC in half which property is this , can u expalin me?
thanks

ABC and ACD are two equilateral triangles, which are mirror images of each other if you join vertices B and D, the segment BD must cut AC, the common base, in half. Which makes BD perpendicular bisector of AC --> so BO is a hight, median, and bisector of angle ABC (O being intersection point of BD and AC).

Hope it's clear.
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Re: If A is the center of the circle shown above  [#permalink]

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This is a proper rhombus... and this is one of the property of rhombus... that they bisect the diagonals in half.. and the diagonals bisect the angles in half...

also Angle (DAC is 120 degree )
and Triangle DAC is an isosceles triangle.. with base BD... hence angle ABD is 30
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Re: If A is the center of the circle shown above  [#permalink]

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thanks bunuel , i also appreciate the explanation given by jivana.
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In triangle ABC, AB = BC = AC meaning all angles = 60 deg each.
Similarly, in Triangle CDA, AD = CD = AC meaning all angles = 60 deg each.

thus, in quadrilateral ABCD, Angle A = 120 = C and Angle B = 60 deg = Angle D.

Thus in Triangle, ABC A + Angle ABD + Angle ADC = 180

120 + 2 * Angle ABD = 180

x = 30 deg.
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Re: If A is the center of the circle shown above (see attachment  [#permalink]

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we are given that AB=BC=CD

This makes ABCD a square. Therefore, BD and AC are the diagonals of the square. Angle A, B, C, D = 90 and hence x=45

Why is this not considered that ABCD is a square?
Attachments Circle.PNG [ 26.39 KiB | Viewed 21556 times ]

Math Expert V
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Re: If A is the center of the circle shown above (see attachment  [#permalink]

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aakrity wrote:
we are given that AB=BC=CD

This makes ABCD a square. Therefore, BD and AC are the diagonals of the square. Angle A, B, C, D = 90 and hence x=45

Why is this not considered that ABCD is a square?

ABCD is not a square it's a rhombus (the diagonals are not equal).
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Re: If A is the center of the circle shown above (see attachment  [#permalink]

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Bunuel wrote:
aakrity wrote:
we are given that AB=BC=CD

This makes ABCD a square. Therefore, BD and AC are the diagonals of the square. Angle A, B, C, D = 90 and hence x=45

Why is this not considered that ABCD is a square?

ABCD is not a square it's a rhombus (the diagonals are not equal).

O right how silly of me I assumed if a quadrilateral has four equal sides, it is a square. Now the solution makes sense to me. Thank you.
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Re: If A is the center of the circle shown above (see attachment  [#permalink]

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Bunuel wrote:
aakrity wrote:
we are given that AB=BC=CD

This makes ABCD a square. Therefore, BD and AC are the diagonals of the square. Angle A, B, C, D = 90 and hence x=45

Why is this not considered that ABCD is a square?

ABCD is not a square it's a rhombus (the diagonals are not equal).

Bunnel,

So Diagonals are 'angle bisectors' both in a Rhombus (not congruent here) and a Square. But it's not the case in a rectangle where they are congruent and perpendicular bosectors alone.
Is my understanding accurate?
I've read that every point on an angle bisector is equidistant from both the adjacent sides. Is there any other way to identify whether a diagonal is also an angle bisector?

Thank you
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Re: If A is the center of the circle shown above (see attachment  [#permalink]

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Kconfused wrote:
Bunuel wrote:
aakrity wrote:
we are given that AB=BC=CD

This makes ABCD a square. Therefore, BD and AC are the diagonals of the square. Angle A, B, C, D = 90 and hence x=45

Why is this not considered that ABCD is a square?

ABCD is not a square it's a rhombus (the diagonals are not equal).

Bunnel,

So Diagonals are 'angle bisectors' both in a Rhombus (not congruent here) and a Square. But it's not the case in a rectangle where they are congruent and perpendicular bosectors alone.
Is my understanding accurate?
I've read that every point on an angle bisector is equidistant from both the adjacent sides. Is there any other way to identify whether a diagonal is also an angle bisector?

Thank you

All is true, except that the diagonals in rectangle are no perpendicular to each other. They bisect each other but not at a right angle.
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GMAT 1: 690 Q50 V34 GMAT 2: 720 Q50 V38 Re: If A is the center of the circle shown above (see attachment  [#permalink]

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Thought BC and CD were the arcs, not straight lines.
Please provide accurate diagrams of clarify in the statement.
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Re: If A is the center of the circle shown above (see attachment  [#permalink]

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fecob wrote:
Thought BC and CD were the arcs, not straight lines.
Please provide accurate diagrams of clarify in the statement.

BC and CD are arcs of the circle AND straight lines as well as in shown in if-a-is-the-center-of-the-circle-shown-above-see-attachment-83638.html#p1049882

There is no need to provide accurate diagrams as the figure has a note: FIGURE NOT DRAWN TO SCALE.
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Re: If A is the center of the circle shown above (see attachment  [#permalink]

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What we know:
1) The figure is a rhombus (Diagonals are angular bisectors)
2) All lines extended from A are radii of the circle.
Therefore, AB = BC = AC. This in turn makes ABC an equilateral triangle. Now you know angleABC is 60*.

As the figure is a rhombus and its diagonals bisect the angleABC, x = 30

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Re: If A is the center of the circle shown above (see attachment  [#permalink]

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When i looked at the question for the first time, i had a thought if BC and CD refers to the length of the Arc and it requires me to use the Arc formula or sth like that..
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Re: If A is the center of the circle shown above (see attachment  [#permalink]

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Triangle abc and triangle adc are both equilateral triangles therefore all their angles are 60 degrees each and the sum of all the 4 x's comes out to be 120 which when we divide by 4 gives us 30 degrees as answer which is option [B] Re: If A is the center of the circle shown above (see attachment   [#permalink] 07 Mar 2018, 09:31

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