How Can We Prove PR = PS = PT = PU in a Non-Rectangular Cyclic Quadrilateral?

[IHateFactorial]

We are given a cyclic quadrilateral, ABCD. ABCD is not rectangular. Lines AD and BC intercept at P. We draw the circumcircles of BDP and ACP. Line AC intercepts the circumcircle of BDP at points S and U, with point U lying between points A and C. Line BD intercept the circumcircle of ACP at points R and T, with point T lying between points B and C. Prove PR = PS = PT = PU.

So far, I've only been able to prove that angles ACB and BDA are equal, which proves that angles ACP and BDP are equal. To prove that PS = PT, I just have to prove that angles TSP and STP are equal, and I'll go from there, but I have no idea what to do...

 

[jvicens]

I would approach this problem using the following steps:

1. Draw a clear and accurate diagram of the given quadrilateral and its circumcircles. Label all the points and angles.

2. Use the given information to identify any congruent angles or sides. In this case, we know that angles ACP and BDP are equal, and we can also see that angles ACB and BDA are equal.

3. Use the properties of cyclic quadrilaterals to find any other congruent angles or sides. For example, opposite angles in a cyclic quadrilateral are supplementary, so we can conclude that angles BCP and ADP are also equal.

4. Look for any other relationships between angles or sides that can help us prove that PR = PS = PT = PU. For example, we can see that angles BCP and ACP are both inscribed angles intercepting the same arc, so they must be equal. Similarly, angles ADP and BDP are both inscribed angles intercepting the same arc, so they must be equal as well.

5. Use these relationships to create congruent triangles. For example, we can create triangle BCP and triangle ACP, which have two pairs of congruent angles (BCP = ACP and BCP = ACP) and a shared side (CP). Therefore, these triangles are congruent by the Angle-Angle-Side (AAS) congruence theorem.

6. Use the congruent triangles to prove that PR = PS = PT = PU. Since triangle BCP and triangle ACP are congruent, their corresponding sides must also be congruent. This means that BP = AP and CP = CP. Therefore, we can conclude that angles BCP and ACP are inscribed angles intercepting the same arc, so their corresponding chords (BP and AP) must also be congruent. This means that PT = PU.

7. Similarly, we can create congruent triangles BDP and ACP, which have two pairs of congruent angles (BDP = ACP and BDP = ACP) and a shared side (DP). Therefore, these triangles are congruent by the Angle-Angle-Side (AAS) congruence theorem. Using the same reasoning as above, we can conclude that PS = PR.

8. Finally, since we have proven that PR = PS = PT = PU, we can conclude that the quadrilateral PRSU is a rhombus