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# If quadrilateral ABCD is inscribed into a circumference

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Senior Manager
Joined: 13 Mar 2007
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29 Aug 2007, 05:32
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E. If quadrilateral ABCD is inscribed into a circumference. What is the value of angle A?

(1) AC=CD
(2) angle D=70 degrees

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Manager
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29 Aug 2007, 06:49
I dont understand what circumference means i am assuming that a quad is inscirbed in a circle.
I think answer should be C

if ac=cd

and if the quadialteral is inscribed in circle then b shud make it a parallelogram. therefore, ab=cd=ac and ang cad =ang cab= ang abc

and ang acd =ang cab

and ang A = ang dac+ ang cab

and we need to know atleast one of these angles
which is provided in stmt 2

hence we need both the stmts

I am not sure whether any other type of quad ( other than llgm ) can be inscribed in the circle with ac=cd

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CEO
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29 Aug 2007, 09:08
E. If quadrilateral ABCD is inscribed into a circumference. What is the value of angle A?

(1) AC=CD
(2) angle D=70 degrees

Don't try this in your head. Draw it on paper. Ok i think i did it right...

A quadrilateral is ANY 4 sided polygon I.E a sqaure, rectangle, rhombus, trapezoid etc...

S1: AC=CD, well ok 2 sides are equal, but what about the other two sides... This doesn't really help us w/ the actual angle. Insuff,

S2: angle D=70degres... This tells us about angle D, but gives us no clue to what angle A is. Insuff.

S1 and S2: To me both of these are still insuff. The 4sided polygon can be drawn in almost an infinite amount of ways, as long as AC=CD and D is 70degrees. However, when drawing the quad. It is easy to see that A could be several different values while satisfying S1 and S2.

Now b/c its inscribed in the circle... Im not sure if I missed something here, or if there is a special rule.

Anyway my ans is E

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Manager
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29 Aug 2007, 12:26
how do u know that ac cd are sides ...not a diagonal and one side...?

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Director
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30 Aug 2007, 22:01
E. If quadrilateral ABCD is inscribed into a circumference. What is the value of angle A?

(1) AC=CD
(2) angle D=70 degrees

Let me try,

I got E

I would add one thing to the definiton blackbelt gave; the sum of the angles of quadrilateral is 360.

1- stm, say two sides are equal, so we can draw an isoceles and the sum of its angles wil be 180 degree. That is BAC+ABC+BCA. So for the second part of the quadrilateral we still have 180 degrees. That is CAD+ADC+ACD, hundreds possibilities...inluding right isoceles.
We need A, here, so as long as BAC or CAD vary we can not find exact value of A. Evethough drawing an asoceles ABC helps us to fix BAC, we can not say the same about CAD, cuz, again, 180 degrees can be distributed among 3 angles in a number of ways.

2-stm, says that D is 70 degrees, this leaves us with 290 degrees for the rest 3 angles, again hundrens of combinations.

Both,
Ok, from 1st we can assume ABC is fixed now, BAC = BCA are fixed angles, leaving us with 180 degrees for the second part,second statement says that CAD + ACD =180 -70=110, again leaving number of possible angles for CAD and ACD. Could be 50 and 60, or vise versa, and that in turn effects A, which is BAC +CAD.

Look, even if we make such assumptions that line AC -diagonal of ABCD is diameter, there are still hundreds of possible A angles.

That was my reasoning...I still am curious about other possible approaches for this problem.

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