hard math problem

Question

In the inscribed quadrilateral PQRS, PQ≅PS≅RS. PQ→ and SR→ meet at an angle with measure 54. Find the measure of the smallest angle in PQRS. The quadrilateral is inscribed in a circle.

Answer 1

So far, I can answer this only conceptually....

To have a quadrilateral inside the circle, the extension of the lines PQ and SR must intersect outside the circle. For starters, make PQ and SR equal. There is now an isoceles triangle with the 54 (given) angle and 2 equal angles of 63 degrees each. total 63+63+54 = 180.

Next: While maintaining the 54 degree angle, make RS longer and PQ shorter. The triangle first becomes right triangle and then becomes obtuse, I think the smallest angle occurs at the point where the circumscribed shape becomes a triangle instead of a quadrilateral--i.e, when segment RQ goes to zero. This, however, violates a stated constraint: "PQ≅PS≅RS"

My conclusion is that this problem is ambiguous and does not have an obvious solution.

Answer 2

Since pq and rs intersect at 54°; the supplementary angle is 126°.