Question

Given square ABCD. Two congruent isosceles triangles, ABP and BCQ, are constructed

with bases AB and BC. The vertex angle for each of these triangles has a measure
of 80°. Point P lies in the interior of the square, while point Q lies outside
of the square. Find the measure of angle BQP.

Answer 1

Answer is 85 and 95!!!!!!!!

Answer 2

85° and 95°

Explanation:

Let's begin noting that a triangle is isosceles if and only if two of its angles are congruent. We can thus find the angle <ABP, recalling that the sum of the interior angles of a triangle is equal to 180°.

Finally, let point K be the intersection between segments BC and PQ, and let's note that the triangle PQB is a right isosceles triangle, since all the angles in a square are equal to 90°, and the two triangles APB and BQC are congruent.

Therefore, the angle BKQ is equal to 180-50-45=85°.

Of course angle BKP=180-85=95°.

Hope this helps :)