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Re: If A is the center of the circle shown above [#permalink]
25 Feb 2012, 11:41

2

This post received KUDOS

Expert's post

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 5504 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.

Re: If A is the center of the circle shown above [#permalink]
27 Feb 2012, 07:42

1

This post received KUDOS

Expert's post

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).

Re: If A is the center of the circle shown above [#permalink]
27 Feb 2012, 07:43

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]
28 Oct 2013, 08:42

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