203. The Equal Sides Rule · equal sides lock equal angles

Try to break it

Drag A up or down on its vertical line, and B left or right on its horizontal line. C mirrors B, so AB = AC at every position. With those two sides equal, the angles at B and C come out equal too. Try to drag the handles so ∠B ≠ ∠C — impossible.

How you build it

Construct an isosceles triangle.

• Mark point B — the left end of the base.
• Mark point C — the right end of the base.
• Draw the base segment BC.
• Construct the perpendicular bisector of BC — it passes through midpoint D.
• Mark the apex A anywhere on the perpendicular bisector — AB and AC come out equal, making triangle ABC isosceles.
• Draw side AB.
• Draw side AC to complete the isosceles triangle.

The proof, step by step

Prove that the base angles of an isosceles triangle are equal.

• Draw AD perpendicular to BC from vertex A.
• In ΔABD and ΔACD: AB = AC (Given), AD = AD (Common), ∠ADB = ∠ADC = 90°.
• ΔABD ≅ ΔACD by RHS Congruence criterion.
• ∠B = ∠C (Corresponding parts of congruent triangles).

Worked example

In ΔABC, AB = AC and ∠B = 55°. Find the measure of ∠A.

Since AB = AC, ∠C = ∠B = 55°. Sum of angles = 180°, so ∠A = 180° - (55° + 55°) = 70°.

• 55°
• 60°
• 70° — correct
• 110°