How do we know chords joining secants are themselves parallel? (Hadamard)

Question

Prove: If we draw two arbitrary secants through the common points $A, B$ of two circles, the chords joining the new intersections of these lines with the two circles are parallel.

Source: Hadamard's Geometry

Below is my proof, to which request verification, feedback, or suggestions.

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Let circles $\bigcirc F, \bigcirc G$ intersect at points $A, B$.

Let lines $a,b$ intersect at some point $X$ (which may be inside, on, or outside either or both circles); intersect $\bigcirc F$ at $A,A_f$ and $B, B_f$ respectively; and intersect $\bigcirc G$ at $A,A_g$ and $B, B_g$.

Then, by Power of the Point, we have $$AX \cdot A_fX = BX \cdot B_fX \\ \frac {A_fX}{B_fX} = \frac {AX}{BX}$$ and likewise $$\frac {A_gX}{B_gX} = \frac {AX}{BX}$$ so $\triangle A_fXB_f \sim \triangle A_gXB_g$ and $A_fB_f \parallel A_gB_g$ as desired.

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See also Does the Power of a Point Theorem apply to a point at infinity?

Answer 1

Here is another way to prove the chords parallel. Join $AB$.

Case I. Secants do not intersect within a circle.

Since $ABHC$ and $ADJB$ are cyclic quadrilaterals, and$$\angle CAB+\angle BAD=\pi$$then their respective supplements$$\angle CHB+\angle DJB=\pi$$and$$CH\parallel DJ$$ Case II. Secants intersect within a circle (figure below).

Since $ABHC$ is a cyclic quadrilateral, and$$\angle CAB+\angle BHC=\angle CAB+\angle BAJ=\pi$$then$$\angle BHC=\angle BAJ=\angle BDJ$$and by alternate interior angles$$CH\parallel DJ$$

Answer 2

[I did not review the proof in detail. Just a comment that we don't have to use the machinery of Power of a Point. OP's usage is essentially only about finding the obvious similar triangles, which we could do so directly.]

For a much simpler approach, we can show that $ \angle XA_f B_f = \angle XBA = \angle XA_gB_g$ by basic circle properties.