Construct a similarity and derive AA

Try

Consider triangle PQR with P=(0,0), Q=(3,0), R=(0,2). Another triangle has P′=(2,1), Q′=(2,7), R′=(−2,1). They look differently oriented, but a concrete sequence matches them: dilate by 2 about the origin, rotate 90 degrees counterclockwise about the origin, then translate by (2,1).

Check all three vertices. The first triangle becomes (0,0),(6,0),(0,4), then (0,0),(0,6),(−4,0), then the three stated target points. This sequence is a similarity witness: an actual transformation argument, stronger than saying that two drawings look proportional.

Predict

Predict the corresponding side ratios. PQ has length 3 and P′Q′ length 6; PR has length 2 and P′R′ length 4. The third pair has lengths √13 and 2√13. Every ratio of image length to original length is 2. The right angle at P moves to P′; the other corresponding angles also agree.

A reflection could reverse orientation without changing these conclusions. Therefore clockwise vertex order need not be preserved by every similarity. What matters is an explicit correspondence, such as P↔P′, Q↔Q′, R↔R′, followed consistently in every side and angle statement.

Experiment

Carry out the sequence on graph paper and record intermediate coordinates rather than only the final triangle. Then reverse it: subtract (2,1), rotate 90 degrees clockwise, and dilate by 1/2. Each operation is invertible, so the witness maps the whole triangle onto the target, not merely three unrelated points.

Original point	After dilation	After rotation	After translation
P	(0,0)	(0,0)	(2,1)
Q	(6,0)	(0,6)	(2,7)
R	(0,4)	(−4,0)	(−2,1)

Draw one original interior point and follow the same sequence. It must be transformed by the same rule; moving individual vertices independently would not constitute a similarity.

Understand

We now derive the AA criterion without assuming it. Let nondegenerate triangles ABC and A′B′C′ have equal corresponding angles at A and A′, and at B and B′. The sides AB and A′B′ have positive lengths. We may use the already established facts that rigid motions preserve distances and angles, positive dilations multiply lengths and preserve angles, and a specified ray direction at a point is unique.

First use a rigid motion to place A on A′ and the ray AB along the ray A′B′. If necessary reflect across that line so that the moved C lies in the same open half-plane as C′. Next dilate about A′ by the positive ratio A′B′/AB. The image of B now lands exactly on B′. Call the image of C the point D.

The angle at A is preserved, so D lies on the ray A′C′: on the chosen side of A′B′ there is only one ray making that specified angle with A′B′. The angle at B is also preserved. With BA mapped to B′A′, D lies on the ray B′C′, by the same one-sided ray uniqueness. These two rays have the unique common point C′ because the triangle is nondegenerate. Thus D=C′, and the constructed similarity maps all three vertices correctly.

This proof does not use “the triangles are similar by AA” midway. It constructs the transformation that establishes similarity. The third angle equality and all three side ratios follow from that witness. Choosing a side of the baseline avoids the reflected ambiguity; reflection is an allowed rigid step, not a reason to reject similarity.

Apply

Given two triangles with two matching angles, write the correspondence first, calculate the scale from a pair of corresponding sides, and describe the alignment and possible reflection. Then explain why the third vertex must land correctly. A numerical angle sum by itself does not supply that transformation argument.

An invalid proof compares AB/A′B′ with BC/A′C′ even though B corresponds to B′ and C to C′. Correct pairs must be matched throughout. Another invalid proof claims that equal angles force equal lengths. Our example has matching angles and a scale of 2; angle data determine shape, not absolute size.

Continue

AA makes later proportionality and right-triangle arguments possible. Its dependencies here are Euclidean rigid-motion and dilation geometry, including the parallel-direction behavior used to preserve angles. The earlier parallel-angle proof makes the role of the Euclidean parallel assumption explicit.

For optional depth, explain why one angle alone leaves freedom: after aligning one vertex and matching one side, the other endpoint can still move along a ray. A second corresponding angle fixes its intersection with a second ray and removes that freedom.