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Two regular pentagons ABCDE and BCFGH have a side BC common. X is the mid point of AE, Y is the mid point of BC and Z is the mid point of GH.

1. If the sides EXA, CYB and GHZ are extended, they will intersect at one point.
2. The point of intersection of these sides will be the center of circle passing through points X, Y and Z.


a. Only 1 Correct
b. 1 and 2 both Correct
c. Nether 1 nor 2 are correct

Hi,
both should be correct ..
But is not a GMAT type Q


Hi Chetan, I agree that the formatting might not be "GMAT-type", still, could you please help me understand any way to prove that both will be true? Maybe with/without help of a diagram. I'm getting a bit lost on this one.
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dixitraghav
Two regular pentagons ABCDE and BCFGH have a side BC common. X is the mid point of AE, Y is the mid point of BC and Z is the mid point of GH.

1. If the sides EXA, CYB and GHZ are extended, they will intersect at one point.
2. The point of intersection of these sides will be the center of circle passing through points X, Y and Z.


a. Only 1 Correct
b. 1 and 2 both Correct
c. Nether 1 nor 2 are correct

Hi,
both should be correct ..
But is not a GMAT type Q


Hi Chetan, I agree that the formatting might not be "GMAT-type", still, could you please help me understand any way to prove that both will be true? Maybe with/without help of a diagram. I'm getting a bit lost on this one.

Hi
The first point can be checked by joining E and H...
The line YC is in the centre of EH, so it becomes altitude.

Second point can be checked by joining the midpoints. Rest can be seen in attached sketch.
Hope it helps.
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IMG_20160130_231408.jpg
IMG_20160130_231408.jpg [ 1.44 MiB | Viewed 3431 times ]

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