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Now, in stmt 1 they are saying that AC (which could be the diameter of the circle) = CD (one of the sides of the quad) - This helps us in understanding that angle ADC = angle DAC = x and that x + x + angle ACD = 180 and that angle ACD = 180 - 2x What we need is angel BAD which is not provided in Stmt 1, thus NS

Stmt 2 gives angel ADC = 70 deg which does not help us in estimating any other angle since a quadrilateral can be a square, parallelogram or a rhombus so angles can be different

Combining stmt 1 and stmt 2 we get angle ADC = 70 deg = angle CAD But we are still missing information about the other angle's i.e. angle ABC + angle BCD + angle ADC + angle BAD = 360 angle ABC + angle BCD + angle 70 + (angle BAC + angle CAD) = 360 angle ABC + angle BCD + angle 70 + (angle BAC + 70) = 360

Rest of the angles above are still missing. Thus, E If we were told what kind of a quad it is then it would be easier to find out the angle.

I think it is important to remember a quadrilateral can be a shape of four sides. Any shape is fine. A warped Trapezium of obtuse angles would be a qdrltlt as well. Am I right? _________________

Explanation: I banged my head against the wall after wasting 15 minutes on this question. I then figured out that I still cannot get to the answer. So it has to be E.

before having a kudos/+1 party please answer the eternal question... "how is ac the diameter?"

I had imagined the quadrilateral ABCD inscribed in the the circle at the very top.... let's say in a semi-circle. now how does AC become the diameter???

Now, in stmt 1 they are saying that AC (which could be the diameter of the circle) = CD (one of the sides of the quad) - This helps us in understanding that angle ADC = angle DAC = x and that x + x + angle ACD = 180 and that angle ACD = 180 - 2x What we need is angel BAD which is not provided in Stmt 1, thus NS

Stmt 2 gives angel ADC = 70 deg which does not help us in estimating any other angle since a quadrilateral can be a square, parallelogram or a rhombus so angles can be different

Combining stmt 1 and stmt 2 we get angle ADC = 70 deg = angle CAD But we are still missing information about the other angle's i.e. angle ABC + angle BCD + angle ADC + angle BAD = 360 angle ABC + angle BCD + angle 70 + (angle BAC + angle CAD) = 360 angle ABC + angle BCD + angle 70 + (angle BAC + 70) = 360

Rest of the angles above are still missing. Thus, E If we were told what kind of a quad it is then it would be easier to find out the angle.

Hey dude! you cant assume AC as the diameter of circle. even if you have assumed this, then the quad will be a Rectangle or Square and its very easy to find the angles with these 2 stmnts.

We know all angles in the the small triangle ACD. But now B could be ANY point above line AC on the cirlce - does it always end up as the same angle?? If so, please elaborate, how to get that angle.

Regarding the question if AC could or could not be the diameter: Statement 2 tells you that it is NOT! Because if AC was the diameter, then ABC and ADC would both be right angles. (but S2 tells you ADC = 70)

We know all angles in the the small triangle ACD. But now B could be ANY point above line AC on the cirlce - does it always end up as the same angle?? If so, please elaborate, how to get that angle.

Regarding the question if AC could or could not be the diameter: Statement 2 tells you that it is NOT! Because if AC was the diameter, then ABC and ADC would both be right angles. (but S2 tells you ADC = 70)

Hey, Theorem : The sum of the opposite angles in cyclic Quadrilateral are supplementary to one another, i.e., their sum is equal to two right angles

S1: isosceles triangle, so Angle ADC=DAC. Not sufficient S2: Angle ADC=70 deg. Not sufficient. Combining we have Angle ACD=40 (Sum of 3 angles in a triangle ADC=180 Deg). Now that alternate opp. Angles are equal AC bisects AB & DC, we can say Angle CAB=40. Therefore Angle BAD=110 deg.

S1: isosceles triangle, so Angle ADC=DAC. Not sufficient S2: Angle ADC=70 deg. Not sufficient. Combining we have Angle ACD=40 (Sum of 3 angles in a triangle ADC=180 Deg). Now that alternate opp. Angles are equal AC bisects AB & DC, we can say Angle CAB=40. Therefore Angle BAD=110 deg.

kudos if you like my explanation !

it is not statex thar it is paralellogram so ur explanation os wrong E

I drew ABCD as a rectangle with the points on the circle. I ended up with answer D and was surprised to find out it was E.

Since Rectangles have 4 rt angles, for statement 1 I drew a diagonal line through the rectangle to form 2 rt triangles, and since ac = cd, I thought it was isosceles thus I had <ADC and < DAC = 45 each therefore able to answer the original question.

For statement 2, Since <ADC=70, I was able to use a similar method and turned the rectangle into 2 rt triangles. I added 90 and 70 and then subtracted from 180 to find <DAC. I then was able to find out the value for <BAD by subtracting <DAC from 90, thus I labeled statement 2 as sufficient. I was wondering what I did wrong since I was off by a lot.