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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 19035 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.

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).

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

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27 Feb 2012, 08:43

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

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

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28 Oct 2013, 09:42

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

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22 Jun 2014, 09:38

Bunuel wrote:

aakrity wrote:

we are given that AB=BC=CD AB is the radius of the circle, so AB=AD

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?

we are given that AB=BC=CD AB is the radius of the circle, so AB=AD

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

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17 Sep 2015, 09:28

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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

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22 Jun 2016, 22:37

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