If A is the center of the circle shown above (see attachment

Question

https://gmatclub.com/forum/download/file.php?id=59139 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 PS21278 Circle - original.PNG 21278-0.png

Bunuel · Accepted Answer

https://gmatclub.com/forum/download/file.php?id=54683 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 PS21278 1.png

Bunuel · Answer

Look at the diagram below: Circle.PNG 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. Answer: B.

Jivana · Answer

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.

Now triangle bad is isosceles, since ab = ad.
Therefore abd = adb = 30, since bad =120.

Hence,  B

MBAhereIcome · Answer

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

pbull78 · Answer

bunuel ,

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

Bunuel · Answer

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.

aakrity · Answer

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?

Bunuel · Answer

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

GMATinsight · Answer

Adding teh video solution to teh thread.

https://www.youtube.com/watch?v=JTUQ_VdoMbc

Batman117 · Answer

Here AB=BC
Where AB is radius.
From figure AC is also radius.
So AB=BC=AC
So ABC is equilateral triangle.
So each side is of 60
So X=60/2
X=30

Posted from my mobile device

BrentGMATPrepNow · Answer

Since all of the radii in a circle have the same length, we know that the 3 red lines below all have the same length. 
 https://i.imgur.com/HRHEhLW.png

Since we are told that AB = BC = CD, we can say that all 5 red lines below have the same length
 https://i.imgur.com/ZS3ODnE.png

This means the two red triangles must be  equilateral triangles , which means all of their angles are 60°
 https://i.imgur.com/gV36uVR.png

Since all 4 sides of the red quadrilateral below have the same length, we know that the quadrilateral is a  rhombus 
 https://i.imgur.com/bVJqjU2.png

In a rhombus, the diagonals bisect the angles. 
So the blue diagonal above, bisects the   60°   angle into two equal angles of 30° each, which means  x = 30

Answer: B

Cheers,
Brent

Rainman91 · Answer

Do we assume that the lines are touching the circle's circumference? Just based on the image?

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If A is the center of the circle shown above (see attachment

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If A is the center of the circle shown above (see attachment

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