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by arifaisal » Sat Nov 12, 2011 6:18 am
In the figure, AB is the diameter of the circle and is parallel to CD. If angle AEB=40 what is angle ACD=?
A)100
B)105
C)115
D)120
E)none
In my books solution, it is written that it can be said that CE=DE but what's the explanation for that..pls explain this fully
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by user123321 » Sat Nov 12, 2011 6:28 am
ACD + DCE = 180
DCE = BAE
but BAE + ABE + 40 = 180
=> BAE = 70 since BAE = ABE
=> ACD = 110

is it E?

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by arifaisal » Sat Nov 12, 2011 6:34 am
thnks man...yes its E
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by user123321 » Sat Nov 12, 2011 6:36 am
arifaisal wrote:it is written that it can be said that CE=DE but what's the explanation for that..pls explain this fully
For this the explanation is. the perpendicular bisector of parallel chords in a circle is same. So obviously any point on it is equidistant from both ends of the chords. That's why CE = DE & AE = BE.

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by subhakam » Mon Mar 04, 2013 5:31 pm
e
user123321 wrote:ACD + DCE = 180
DCE = BAE
but BAE + ABE + 40 = 180
=> BAE = 70 since BAE = ABE
=> ACD = 110

is it E?

user123321
STILL DO NOT understand how this happens
but BAE + ABE + 40 = 180
=> BAE = 70 since BAE = ABE
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by GMATGuruNY » Mon Mar 04, 2013 9:51 pm
subhakam wrote: STILL DO NOT understand how this happens
but BAE + ABE + 40 = 180
=> BAE = 70 since BAE = ABE
AB || CD.
Parallel chords intercept CONGRUENT ARCS.
Thus, arc AC = arc DB.
Congruent arcs have CONGRUENT CHORDS.
Thus, in trapezoid ACDB, side AC = side DB.
Thus, trapezoid ACDB is ISOSCELES, implying that ∠BAC = ∠ABD.
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