When two sides and an angle are not enough

Now change one thing. Take ∆ABC and ∆XYZ with AB = XY = 6 cm, AC = XZ = 4 cm, and ∠B = ∠Y = 30°. The angle is still equal — but it is no longer BETWEEN the two equal sides. Are the triangles congruent?

Ask the same question: can non-congruent triangles exist with these measurements? Construct and find out. A rough diagram first helps you plan.

Step 1: draw the base PQ of length 6 cm. Step 2: draw a line l from P making an angle of 30° with PQ. Step 3: draw a sufficiently long arc from Q of radius 4 cm, cutting the line l.

Now look carefully at what the arc does. It crosses the line l at TWO different points — call them R and S.

So which is the third vertex? Both. ∆PQR satisfies all the given measurements, and so does ∆PQS. Both are legitimate answers to the same question.

And unlike Rabia's two triangles in the SSS case, these two are NOT mirror images of each other — they lie on the same side of the base and have visibly different shapes. So we can draw two non-congruent triangles from these measurements.

This is called the SSA — Side Side Angle — condition, and we have just shown that it does NOT guarantee congruence. That is why the word 'included' matters so much in SAS: move the angle out from between the two sides and the guarantee is lost.