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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

Sagot :

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°.

[tex] \angle \: ABP = \frac{180 - 80}{2} = 50 \degree[/tex]

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 :)

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