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GMAT CLUB TEST m18 - question 15

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GMAT CLUB TEST m18 - question 15 [#permalink] New post 22 Jul 2010, 16:14
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In rectangle ABCD ( BC >AB] ), E is the point of intersection of diagonals. If angle Angle ABD is twice the angle EAD , what is the value of angle CED ?
* 30 degrees
* 45 degrees
* 60 degrees
* 90 degrees
* 120 degrees

angle EAD = angle BDA . angle ABD + angle BDA + 90 = 2*angle BDA + angle BDA + 90 = 180.

From this equation, angle BDA = 30 . angle BDC = 90 - angle BDA = 60 . Thus, triangle CED is equilateral and angle CED = 60

The correct answer is C.

Can someone explain how angle EAD = angle BDA as mentioned in the solution?
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Re: GMAT CLUB TEST m18 - question 15 [#permalink] New post 23 Jul 2010, 01:09
Hi,

Diagonal of a rectangle are equal and bisect each other.
Thus EA=ED and angle EAD = angle EDA = angle BDA

regards,
Jack
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Re: GMAT CLUB TEST m18 - question 15 [#permalink] New post 23 Jul 2010, 04:53
can someone post a graph maybe explaining the relations?

The forumulas have me totally confused, as I keep mixing up the angels and lose track of where which angle was in what relation...

Thanks!
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Re: GMAT CLUB TEST m18 - question 15 [#permalink] New post 23 Jul 2010, 11:23
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Hi,

Consider the rectangle ABCD (BC>AD) in the doc attached.
Given facts: angle ABD = 2 * angle EAD
Since diagonals of a rectangle are equal and bisect each other, we have EA=ED. Implying angle EAD = EDA.
Lets consider angle EAD = angle EDA = x.
We already know all angles in a rectangle are 90. Therefore angle BAD = 90
Considering the triangle BAD, sum of angles in a triangle is 180.
So, angle BAD + angle ABD + angle BDA = 180
90 + 2x + x = 180 => 3x = 90 => X = 30
Now, angle EDC = 90-30 = 60
Consider triangle ACD, again angle CAD + angle ADC + angle DCA = 180
30 + 90 +angle DCA = 180 = > angle DCA =60
Knowing 2 angles of triangle EDC to be 60, makes it a equilateral triangle, hence angle CED =60.

Hope its not confusing!
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Last edited by WithHopesToWin on 23 Jul 2010, 11:25, edited 1 time in total.
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Re: GMAT CLUB TEST m18 - question 15 [#permalink] New post 23 Jul 2010, 11:24
btw, thanks Jakolik!!
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Re: GMAT CLUB TEST m18 - question 15 [#permalink] New post 23 Jul 2010, 22:15
WithHopesToWin wrote:
btw, thanks Jakolik!!


You're welcome :)
Re: GMAT CLUB TEST m18 - question 15   [#permalink] 23 Jul 2010, 22:15
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