Prove that the four points are concyclic if they are harmonically related w.r.t. the midpoint of connecting line segment.

$$(1)AB$$ and $$CD$$ are two intersecting line segments, and $$P$$,$$Q$$ are their respective midpoints. If $$AB$$ bisects $$\angle CPD$$ and $$PA^2=PB^2=PC.PD$$, then prove that the points $$A,B,C$$ and $$D$$ are concyclic.

I have a proof of the converse of the above theorem. Consider four concyclic points $$A(a),B(b),C(c)$$ and $$D(d)$$. Let $$O$$ be the midpoint of $$CD$$ . Then the cross ratio of these points is, $$\lambda=\frac {AC.BD}{AD.BC}$$ $$= \frac {(a-c)(b-d)}{(a-d)(b-c)}=-1$$ Rearranging we get: $$(a+b)(c+d)=2(ab+cd)$$ Rearranging again, we get: $$\left\{a-\frac {1}{2}(c+d)\right\}\left\{b-\frac {1}{2}(c+d)\right\}=\left\{\frac {1}{2}(c-d)\right\}^2$$ This implies: $$OA.OB=OC^2=OD^2$$ That is,$$OA$$and $$OB$$ are equally inclined to $$CD$$. $${ }$$ I could have written all of these lines in reverse, but I don't think that would be just, for both the theorems are converses of each other. Is there a distinct proof of theorem $$(1)$$ either using Euclidian geometry or complex numbers (other than this way)? Any help would be appreciated.

Let $$CD\cap CD=\{E\}$$, $$E$$ placed between $$P$$ and $$B$$ and between $$C$$ and $$Q$$.

Thus, since $$PE$$ is a bisector of $$\Delta PCD$$, we obtain: $$PE^2=PC\cdot PD-CE\cdot ED=PA^2-CE\cdot ED.$$ Id est, $$CE\cdot ED=PA^2-PE^2=(PA-PE)(PA+PE)=BE\cdot AE,$$ which gives $$\frac{CE}{AE}=\frac{EB}{ED},$$ which says $$\Delta CEB\sim\Delta AED,$$ which gives $$\measuredangle BCD=\measuredangle BAD$$ and quadrilateral $$ACBD$$ is cyclic.

$$\frac{PC}{PB}=\frac{PB}{PD}, \ \angle CPB=\angle BPD \implies \triangle CPB\sim \triangle BPD$$

These triangles have the same angles ($$\alpha,\beta,\gamma$$) as shown.

$$\frac{PC}{PA}=\frac{PA}{PD}, \ \angle CPA=\angle APD \implies \triangle CPA\sim \triangle APD$$

These triangles have the same angles ($$\delta,\varphi,\varepsilon$$) as shown.

In quadrilateral $$ABCD:$$

$$\angle C= \varphi+\gamma$$

$$\angle D= \beta+\varepsilon$$

$$\angle C +\angle D=\varphi+\gamma + \beta+\varepsilon=180^\circ$$

The last is obvious from triangle $$ABC$$. Sum of opposite angles is $$180^\circ$$ and therefore the quadrilateral is concyclic.