A Forgotten Similarity Theorem?

Exploration

Construction: Follow the step by step construction described below:
Step 1: Construct a dynamic △ABC and a dynamic line segment QR.
Step 2: Measure the ratios AB/BC and AC/BC.
Step 3: Mark ratio AB/BC as a ‘Scale Factor’, and mark Q as a ‘Center’, then ‘Dilate’ the point R from Q as centre with the marked ratio AB/BC. With Q as centre draw a circle with radius QR' where R' is the image of the preceding dilation.
Step 4: Mark ratio AC/BC as a ‘Scale Factor’, and mark R as a ‘Center’, then ‘Dilate’ the point Q from R as centre with the marked ratio AC/BC. With R as centre draw a circle with radius Q'R where Q' is the image of the preceding dilation.
Step 5: Construct the intersection of the two circles, and label one of the intersections as P. (We ignore the other intersection, say P', since △P'QR is congruent to △PQR).

Explore: In the interactive sketch below, Steps 1-2 have already been done. Click on the 'Show Steps 3-5' buttons to display the next steps.
1) What do you notice about triangles ABC and PQR? Check your observation by dragging any of the red points.
2) Can you formulate your findings as a theorem?

Forgotten Similarity Theorem

A Forgotten Similarity Theorem
If for two triangles ABC and PQR any two pairs of ratios hold from the following three pairs of corresponding ratios, AB/BC = PQ/QR, BC/AC = QR/PR and AB/AC = PQ/PR, then △ABC is similar to △PQR.

Challenge
Can you logically explain why (prove that) the above theorem is true?

Note
Note that the theorem above is equivalent to the SSS-similarity theorem - it implies the SSS-similarity theorem and is implied by it. Perhaps that is the reason why it is not mentioned in geometry texts?

Reference
A paper A Forgotten Theorem for Triangle Similarity? with a proof of the above result has been published in the Learning & Teaching Mathematics (LTM) journal, Dec 2022, no. 33, published by the Association for Mathematics Education of South Africa.