Chapter 7.4 — Exercise 7.3 — More Congruence Criteria

SSS: When Three Sides Settle Everything

SSS is in some ways the most intuitive of the four rules: build a triangle out of three fixed-length rods hinged at their ends, and there's only one possible shape it can take — no amount of pushing or pulling changes any of its angles once all three side lengths are locked in. RHS looks like a special, restricted case of SAS at first glance — after all, it's also fixing two sides and an angle — but the angle it fixes is a non-included one (the right angle sits opposite the hypotenuse, not between the two known sides), which is exactly the SSA arrangement that fails to guarantee congruence in general. RHS is the single narrow exception where a non-included angle does still work, and it works only because that angle is locked at exactly 90°.

SSS Feeding Directly Into SAS

Triangle ABC has sides AB and BC and median AM equal respectively to sides PQ, QR and median PN of triangle PQR; the goal is to show first △ABM ≅ △PQN, then the full △ABC ≅ △PQR. Since M and N are midpoints of BC and QR, BM = ½BC and QN = ½QR — and because BC = QR is given, BM = QN follows immediately. In △ABM and △PQN: AB = PQ (given), AM = PN (given, the medians), and BM = QN (just shown) — three sides, SSS, so △ABM ≅ △PQN.

That first congruence hands over a fresh angle equality by CPCT: ∠ABM = ∠PQN, which is the same angle as ∠ABC = ∠PQR. Now, in △ABC and △PQR: AB = PQ (given), BC = QR (given), and the angle between them, ∠ABC = ∠PQR (just derived) — SAS, so △ABC ≅ △PQR. The whole proof is really two separate congruences stacked in sequence, where the first one's CPCT output becomes the second one's third given fact — a pattern worth watching for whenever a problem hands over more matching parts than a single rule seems to need at once.

RHS Proves a Triangle Isosceles From Its Altitudes

BE and CF are two equal altitudes of triangle ABC, and the goal is to prove the triangle isosceles using RHS specifically. In △BEC and △CFB: ∠BEC = ∠CFB = 90° (both altitudes), BE = CF (given), and BC = CB — the shared hypotenuse of both right triangles. Hypotenuse and one side matching, RHS, so △BEC ≅ △CFB, and CPCT gives ∠ECB = ∠FBC — which is the same statement as ∠C = ∠B in the original triangle. Equal angles force AB = AC by the converse fact from the previous exercise, so △ABC is isosceles.

Notice that the shared side BC is playing the hypotenuse role in both right triangles at once, which is exactly why this pairing works for RHS specifically and wouldn't work as cleanly if the two altitudes were compared using AAS instead — the shared side needs to be doing hypotenuse duty on both sides for RHS's own conditions to actually be met.

The Isosceles Base-Angle Theorem, Proved a Second Way

The previous exercise proved that AB = AC forces ∠B = ∠C using SAS on a bisected angle at A. This exercise proves the identical result a second time, using RHS instead: draw AP ⊥ BC. In △APB and △APC: ∠APB = ∠APC = 90° (construction), AB = AC (given), and AP = AP (shared) — hypotenuse and one side, RHS, so △APB ≅ △APC, and CPCT gives ∠B = ∠C.

Having two independent proofs of the exact same theorem is a genuinely useful thing to sit with, not just a curiosity: it means the isosceles base-angle result doesn't depend on which construction happens to be drawn first, angle bisector or perpendicular altitude — both constructions land on the same point when AB = AC, a coincidence that's really no coincidence at all once triangle symmetry is taken into account.

An Angle-Chasing Proof Using External Angles

ABC is isosceles with AB = AC, and side BA is extended past A to a point D with AD = AB; the goal is to show ∠BCD is a right angle. Calling ∠ABC = ∠ACB = x° (base angles of the isosceles triangle), the exterior angle ∠DAC works out to 2x° since it's the exterior angle of triangle ABC at A, equal to the sum of the two remote interior angles x° + x°. Meanwhile triangle ADC is itself isosceles (AD = AC, since AD = AB = AC), so ∠ADC = ∠ACD = y° for some y°, and by the same exterior-angle reasoning applied to triangle ADC, ∠BAC = 2y°. Since ∠DAC and ∠BAC together form a straight line at A, ∠DAC + ∠BAC = 180°, giving 2x° + 2y° = 180°, so x° + y° = 90° — and x° + y° is exactly ∠BCD, since it's built from ∠ACB and ∠ACD placed side by side. So ∠BCD = 90°.

This proof leans on the exterior-angle property rather than on any of the four congruence rules at all, which makes it a useful reminder that congruence isn't the only tool this chapter has assembled — angle-chasing through isosceles triangles and straight lines can settle a result just as completely on its own.

Two Quick Closing Results

A right-angled isosceles triangle with ∠A = 90° and AB = AC reaches ∠B = ∠C by drawing AD ⊥ BC and applying RHS to △ADB and △ADC exactly as before — a specific case of the base-angle theorem rather than a new idea. Separately, an equilateral triangle's three equal sides give ∠B = ∠C (from AB = AC) and ∠B = ∠A (from BC = CA) simultaneously, so all three angles are equal; since they sum to 180°, each one must be exactly 60°.

A related figure extends the isosceles triangle ABC (AB = AC) by producing BA past A to Q and CA past A to P, with AQ = AP, and asks for PB = QC. In △ABP and △ACQ: AB = AC (given), ∠PAB = ∠QAC (vertically opposite, since P, A, C and Q, A, B are each straight lines crossing at A), and AP = AQ (given) — SAS once again, so △ABP ≅ △ACQ, and CPCT gives PB = QC directly. This closing problem is really a callback to the very first move taught in Exercise 7.1: spotting a vertically opposite angle pair as the "free" third fact that completes an SAS case, now applied to a figure built from an isosceles triangle rather than an arbitrary quadrilateral.

Toward the Chapter's Final Rules About Inequality

All four congruence rules — SAS, ASA, SSS, and RHS — are now available. Exercise 7.4 shifts away from equality entirely, asking instead how a triangle's side lengths and angle measures constrain each other through inequalities rather than exact matches.