126. SAS Congruence Criterion · Two sides and the included angle lock the shape

Try to break it

Drag D anywhere. The second triangle moves as a rigid block — two sides and the included angle stay locked to match the first triangle. Try to distort its shape; the SAS condition won't let you.

How you build it

Construct triangle ABC from a side AB, the included angle ∠A, and a second side AC.

• Draw segment AB. This is the first of the two given sides — call its length c.
• At vertex A, use the angle tool to construct the given included angle ∠A. This draws a ray from A — vertex C will lie somewhere along this ray.
• With centre A and radius b (the second given side), draw an arc that cuts the new ray. Mark the intersection as C. Now AC = b exactly.
• Draw segment BC to close triangle ABC. With side AB = c, included angle ∠A, and side AC = b all fixed, the third side BC has only one possible length — by SAS, any triangle built from these three pieces is congruent to this one.

The proof, step by step

Prove that two sides and the included angle make the triangles congruent (SAS).

• Given: AB = DE, AC = DF, and ∠BAC = ∠EDF.
• Place ΔDEF on ΔABC so that D coincides with A and DE lies along AB.
• Since DE = AB, point E must coincide with point B.
• Since ∠EDF = ∠BAC, the ray DF must lie along the ray AC.
• Since DF = AC, point F must coincide with point C.
• Therefore, EF coincides with BC, and ΔABC ≅ ΔDEF.

Worked example

In ΔABC and ΔDEF, AB = DE, AC = DF, and ∠A = ∠D. Which congruency test proves that ΔABC ≅ ΔDEF?

Since two sides and the included angle of one triangle are equal to the corresponding two sides and included angle of the other triangle, the SAS (Side-Angle-Side) Congruency Test applies.

• SSS Test
• SAS Test — correct
• ASA Test
• RHS Test