Cyclic Quadrilateral with an equilateral triangle as a part of it

Question

Find the relation between RX, RY and RZ if XYZ is an equilateral triangle.

Answer 1

Explanation:
Let the intersection of XR and YZ be O.
∠XRY = ∠XRZ

∠ORY = ∠XRZ
∠RYZ = ∠RXZ
Therefore, Triangle YOR is similar to Triangle XZR.
=> XZ.YR = OY.XR --(1)

∠XRY = ∠ORZ
∠YXR = ∠YZR
Therefore, Triangle ZOR is similar to Triangle XYR.
=> XY.ZR = OZ.XR --(2)

Adding (1) and (2)
XZ.YR + XY.ZR = OY.XR + OZ.XR
=> XY(YR + ZR) = XR.YZ
=> RY + RZ = RX
Thus Proved.

Answer 2

Ptolemy's Theorem says that $RX.YZ=RY.XZ+RZ.YX$.

Since $XY=YZ=ZX$, this simplifies to $RX=RY+RZ$.

Answer 3

Hint: Use the cosine rule for the triangles XYR and XZR to establish relationship among RX, RY and RZ. (keep in mind ∠XRY = ∠XRZ)

Answer 4

Answer not using Ptolemy

Angle $XRZ=60^o$ by angle on a chord. So, by the cosine rule:

$XZ^2=XR^2+RZ^2-XR.RZ$ and, similarly

$XY^2=XR^2+RY^2-XR.RY$.

Subtracting

$RZ^2-XR.RZ=RY^2-XR.RY$

$RZ^2-RY^2=XR.RZ-RX.RY$

$(RZ-RY)(RZ+RY)=RX(RZ-RY)$

So $RX=RY+RZ$ or $RY=RZ$.

The case $RY=RZ$ produces a simple configuration which also gives $RX=RY+RZ$.