Vlad77 wrote:

E. If quadrilateral ABCD is inscribed into a circumference. What is the value of angle A?

(1) AC=CD

(2) angle D=70 degrees

Please, explanations ONLY

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.

Please, draw. And you will see that A=BAC+CAD; C= BCA+ACD.

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.