Exercise 9. Let $A B C$ be an isosceles triangle at $A$ such that $A B < B C$. Let $P$ be the point on the line $(B C)$ located outside the segment $[B C]$ such that $B P = B A$. Let $Q$ be the point on $(A P)$ different from $A$ such that $C Q = C A$. If $\widehat{Q P C} = x$, what is the measure of the angle $\widehat{P C Q}$?
## Solution to Exercise 9

![](https://cdn.mathpix.com/cropped/2024_04_29_ebc2d68ed68a9c93df18g-14.jpg?height=571&width=1294&top_left_y=411&top_left_x=381)

In this exercise, it is necessary to write down all the equations that translate the assumptions of the problem statement.

Here, the triangles \( Q C A, B A C \) and \( P B Q \) are isosceles, so the base angles are equal, which gives \(\widehat{C Q A} = \widehat{C A Q}, \widehat{A B C} = \widehat{A C B}\) and \(\widehat{B A P} = \widehat{B P A}\).

We start by decomposing the angle \(\widehat{P C Q}\) into the angles \(\widehat{Q C A}\) and \(\widehat{A C B}\), and we express these angles separately in terms of \(x\).

On one hand, since the points \(C, B\) and \(P\) are collinear, we have:

\[
\widehat{A C B} = \widehat{A B C} = 180^{\circ} - \widehat{A B P} = \widehat{B A P} + \widehat{B P A} = 2x
\]

On the other hand, since the points \(P, A\) and \(Q\) are collinear and the sum of the angles in the triangle \(B A C\) is \(180^{\circ}\), we have:

\[
\widehat{Q A C} = 180^{\circ} - \widehat{B A C} - \widehat{B A P} = \widehat{A C B} + \widehat{A B C} - \widehat{B A P} = 2x + 2x - x = 3x
\]

We can then calculate the angle \(\widehat{Q C A}\) in terms of the angle \(x\), by considering the triangle \(Q C A\):

\[
\widehat{Q C A} = 180^{\circ} - \widehat{C Q A} - \widehat{Q A C} = 180^{\circ} - 2 \widehat{Q A C} = 180^{\circ} - 6x
\]

Finally, we can conclude:

\[
\widehat{P C Q} = \widehat{Q C A} + \widehat{A C P} = 180^{\circ} - 6x + 2x = 180^{\circ} - 4x
\]