Geometric Twins · Lesson 6 of 10
Two Angles and a Side
“Use two angles and one length to fix a triangle, and connect angle facts with congruence proofs.”
• Construct a triangle from two angles and their included side. • Apply the ASA condition using corresponding parts. • Derive AAS from the triangle angle sum and ASA. • Use midpoint and vertically opposite angle facts in a congruence proof.
Two rays fix the third vertex
Three angles alone leave a triangle’s size free. Adding a specified side removes that freedom. Suppose BC = 5 cm, angle ABC = 50°, and angle ACB = 30°. The side BC lies between the two given angle vertices, so it is their included side. Once BC is drawn, the two angles locate rays from its endpoints, and those rays meet at A.
Draw BC = 5 cm. At B construct a 50° ray, and at C construct a 30° ray, choosing both rays on the same side of BC. Their intersection is A. A second triangle made from these same measurements fits this one because the fixed base and fixed directions determine the same third vertex. Constructing on the opposite side gives a congruent reflected copy.
Two triangles are congruent if two corresponding angles and the side included between those angles are equal. ASA stands for Angle–Side–Angle.
Problem
BC = YZ = 5 cm, angle B = angle Y = 50°, and angle C = angle Z = 30°. What follows?
- 1.BC connects the two given angle vertices B and C. YZ connects their matches Y and Z.
- 2.Both angle pairs and their included side pair agree, so A ↔ X, B ↔ Y, and C ↔ Z.
- 3.Triangle ABC ≅ triangle XYZ by ASA. The third angles both equal 180° − 50° − 30° = 100°. The other side pairs also agree.
A side outside the two angle vertices
What if the given side is not between the two given angles? Consider angle A = 35°, angle C = 75°, and BC = 4 cm. BC is not the side connecting A and C, so the information is called AAS: Angle–Angle–Side. It still fixes the triangle because the two angles determine the third angle.
The angles of a triangle add to 180°. Therefore angle B = 180° − 35° − 75° = 70°. If a second triangle has the same given angles, its third angle is also 70°. Now BC lies between the known angles B and C, so the information becomes ASA. This is why AAS is a valid condition as well.
Two triangles are congruent if two corresponding angles and a corresponding side not included between those angles are equal. It follows from the angle sum and ASA.
Problem
Angle A = angle X = 35°, angle C = angle Z = 75°, and BC = YZ = 4 cm. Prove congruence.
- 1.Angle B = 180° − 35° − 75° = 70°. The same subtraction gives angle Y = 70°.
- 2.We now have angles B = Y and C = Z, with their included sides BC = YZ.
- 3.By ASA, ABC ≅ XYZ. The original information was AAS, so it is equally correct to state the AAS condition.
Match the given side as well as the angles. Two angles alone do not fix size; adding any arbitrary equal side without checking its correspondence is not enough.
Find useful facts before choosing a condition
In a diagram, equal parts may follow from familiar facts rather than be given as measurements. A midpoint splits a segment into two equal lengths. Vertically opposite angles are the opposite angles formed by two intersecting straight lines, and they are equal. Combining these facts can supply a complete congruence condition.
Problem
O is the midpoint of both AD and BC. Show that AB = DC.
- 1.Because O is the midpoint of AD, AO = OD. Because O is the midpoint of BC, BO = OC.
- 2.Angles AOB and DOC are vertically opposite, so they are equal. These angles lie between the two equal segment pairs.
- 3.Thus AOB ≅ DOC by SAS, with A ↔ D, O ↔ O, and B ↔ C.
- 4.The corresponding outer sides are AB and DC. Therefore AB = DC. Although this lesson introduces ASA and AAS, SAS is the condition supplied by these particular facts.
A proof has two stages: establish congruence using three sufficient facts, then read the required equality from the correspondence. Naming the condition must follow the information actually available. Not every problem in a lesson on angles uses ASA; the midpoint example above uses SAS.
Quiz
Which side is included between angles B and C?
Two angles are 50° and 30°. What is the third angle?
Why is AAS sufficient?
Which facts establish ASA?
In the midpoint example, which statement gives the correct correspondence?
Practice Problems
- Construct BC = 5 cm, angle B = 50°, and angle C = 30°. Locate A and calculate angle A.
- Explain why giving the same three angles without BC would not fix the size.
- If angle P = angle L = 45°, angle R = angle N = 65°, and QR = MN, prove PQR ≅ LMN.
- Draw intersecting AD and BC with O the midpoint of each. Prove AOB ≅ DOC and then AB = DC.
- Classify angle A = D, angle C = F, AC = DF as ASA or AAS and explain your choice.
- Classify angle A = D, angle C = F, BC = EF as ASA or AAS, and calculate the third angle if A = 35° and C = 75°.
Key Takeaways
• ASA gives two angles and the side joining their vertices. • AAS gives two angles and another corresponding side. • The angle sum converts AAS information into ASA. • A measured side fixes size that angle information alone leaves free. • Midpoints, shared sides, and vertically opposite angles can supply facts for proofs.