Chapter 7.2 — Exercise 7.1 — Congruency Problems

A Shared Side Completes the Triangle

Quadrilateral ACBD has AC = AD, and AB bisects ∠A. In △ABC and △ABD: AC = AD (given), ∠CAB = ∠DAB (AB bisects the angle), and AB = AB — the same segment, shared by both triangles. Three facts, SAS satisfied, so △ABC ≅ △ABD, and CPCT immediately gives BC = BD.

The shared side AB is doing something subtly different from the other two facts: it isn't stated as given information about the figure, it's simply true because both triangles happen to include that exact same segment. Spotting a shared side or shared angle this way — "common side" or "common angle" as a justification — is one of the most frequent moves across this whole exercise. The genuine difficulty in most of these problems isn't the congruence rule itself, which is usually easy enough to spot once the two relevant triangles are correctly identified — it's spotting which two triangles, out of everything drawn in a busier figure, are actually the useful pair worth focusing the whole proof on in the first place.

Vertical Angles Complete an AAS Case

AD and BC are equal, both perpendicular to segment AB, meeting a third line at point O. In △OAD and △OBC: AD = BC (given), ∠AOD = ∠BOC (vertically opposite), and ∠OAD = ∠OBC = 90° (given) — two angles and a non-included side, AAS. So △OAD ≅ △OBC, giving OA = OB by CPCT — meaning CD bisects AB exactly at O.

This is a useful contrast with the SAS case above: here it's an angle relationship (vertically opposite angles, guaranteed whenever two lines cross) supplying the "free" third fact, rather than a shared side. Recognising which kind of free fact a figure offers — a common side, a common angle, or a vertically-opposite pair — is most of the work in choosing the right rule. None of these three "free" facts ever needs to be stated as given information in the problem, which is exactly why they're easy to overlook on a first read-through: a shared side is true simply because two triangles happen to share an edge, and vertically opposite angles are true purely because two lines happen to cross, regardless of anything else the problem specifies. Notice, too, that AAS and ASA are not the same rule wearing two different names: ASA fixes the side sitting between the two known angles, while AAS fixes a side that sits outside them, adjacent to just one. Both still pin the triangle down completely, because knowing any two angles of a triangle already fixes the third by subtraction from 180° — so AAS quietly reduces to ASA the moment that third angle is worked out, even though the two given angles on paper look positioned differently at first glance.

Congruent Right Triangles via RHS

D is the midpoint of BC in △ABC, with DE ⊥ AB, DF ⊥ AC, and DE = DF given. In △BED and △CFD: DE = DF (given), ∠DEB = ∠DFC = 90° (given), and BD = CD (D is the midpoint) — hypotenuse and one side matching in two right triangles, which is exactly the RHS rule. △BED ≅ △CFD follows directly. RHS is the only one of the four congruence rules that specifically requires a right angle to already be present in both triangles, which is exactly why it's reached for the moment two perpendiculars show up together in a proof like this one, rather than trying to force the same conclusion out of SAS or AAS instead.

A Construction That Proves a Triangle Isosceles

If the bisector of ∠A in △ABC also bisects BC at D, the triangle turns out isosceles — but proving it needs a construction first: extend AD to a point E with DE = AD, then join C to E. In △ABD and △EDC: AD = ED (construction), ∠ADB = ∠EDC (vertically opposite), and BD = CD (given) — SAS gives △ABD ≅ △EDC, so AB = EC and ∠BAD = ∠CED by CPCT. Since ∠BAD = ∠CAD was already given, ∠CAD = ∠CED, which makes △ACE isosceles too (angles opposite equal sides), so AC = EC. Combining AB = EC and AC = EC gives AB = AC directly — △ABC is isosceles.

Constructing a new point is a genuinely different move from the SAS/AAS/RHS problems above: none of the original figure's triangles were directly congruent to each other, so the proof builds an auxiliary triangle specifically to create one that is, then routes the answer back through it. Choosing exactly where to place the new point isn't arbitrary, either, even though it might look that way at first: E is defined specifically so that D — the already-established midpoint of BC — also becomes the midpoint of AE, which is precisely the shared condition that makes △ABD and △EDC provably congruent by SAS in the first place, rather than just two vaguely similar-looking triangles sitting near each other in the figure. It's also worth noticing what the construction step is quietly assuming is even possible: any ray can always be extended past a point by any chosen length, so producing AD to E with DE = AD is never in question — the only genuine work is deciding where, specifically, to stop, and here that stopping point is chosen precisely to hand the proof a second triangle sharing D as a common midpoint with the first.

Reading Backward From an Angle Condition

In right triangle ABC with the right angle at B, ∠BCA = 2∠BAC is given, and the goal is to show AC = 2BC. Producing CB to a point D with BD = BC, then joining A to D, gives △ABD ≅ △ABC by SAS (BD = BC, ∠ABD = ∠ABC = 90°, AB shared) — so AD = AC and ∠BAD = ∠BAC by CPCT. Calling ∠BAC = x°, angle-chasing shows ∠DAC = 2x° = ∠BCA, which makes △ADC isosceles with AD = CD. Since AD = AC already, CD = AC too, and AC = BC + BD = BC + BC = 2BC. The whole chain hinges on one geometric fact that's easy to state but easy to rush past: because D was constructed so that B sits exactly halfway between C and D, the full length CD splits cleanly into CB plus BD, and since those two pieces are equal by construction, CD collapses to simply twice BC — the final step the entire argument was really building toward from the very first line.

From Two Rules to Four

Every proof here has leaned on SAS, ASA, or RHS. Exercise 7.2 turns one of these techniques into a named theorem about isosceles triangles specifically, and Exercise 7.3 adds the two remaining rules, SSS and a formal statement of RHS, completing the full four-rule toolkit.