Will award brainliest to correct answer! Given square ABCD. Two congruent i s o - Δ's ABP and BCQ are constructed with bases AB and BC. Each of these triangles has vertex angle of 80°. Point P lies in the interior of the square, while point Q lies outside of the square. Find the angle measure between PQ and BC.

Thanks in advance :D

85° and 95°

Step-by-step 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 :)

1     sepan

2     encuentren

3     tenga

4     hable

5   parezca

6   llegue

the correct answer is a) "catholic church".

the proposals that liberals in mexico made during 1855 were particularly focused on a person's individual rights, such as the freedom of speech, the freedom of conscience, freedom of assembly, and the right to possess weapons.

when the constitution was formally published 2 years later (1857). it also contained an article which permitted education free from any religious influence, the removal of the catholic church institutional privileges, and the sale of property belonging to the church. all of these provisions were clearly against the catholic church interests.