Chapter 7.3 — Exercise 7.2 — Properties of Triangles

Proving It: One Bisector, Two Congruent Halves

Start with isosceles △ABC, AB = AC, and draw AD as the bisector of ∠A, meeting BC at D. In △ABD and △ACD: AB = AC (given), ∠BAD = ∠CAD (AD bisects ∠A by construction), and AD = AD — shared by both triangles. Three matching facts, two sides and the included angle, so SAS applies directly: △ABD ≅ △ACD. CPCT then hands over exactly the result being chased: ∠ABD = ∠ACD, which is the same statement as ∠B = ∠C in the original triangle.

The construction is doing something worth noticing on its own: nothing in the original triangle came with a bisector already drawn, so the proof manufactures one purely as a tool, uses the congruence it creates, then reads the needed angle equality straight off CPCT. This is the same auxiliary-construction move seen in the trickier problems of the previous exercise, just applied here to prove a general theorem rather than to solve one specific numbered figure. A genuinely useful companion fact travels alongside this theorem, and it runs in the exact opposite direction: if two angles of a triangle are already known to be equal, then the sides sitting opposite those two angles must themselves be equal too — a converse statement that turns out to be just as reliable as the original theorem, and gets used just as often across the problems below.

Angle Bisectors Meeting Inside the Triangle

In isosceles △ABC with AB = AC, the bisectors of ∠B and ∠C meet at a point O; joining A to O, the goal is to show OB = OC and that AO bisects ∠A. Since AB = AC, ∠B = ∠C by the theorem above, and halving equal angles keeps them equal, so ∠OBC = ∠OCB — meaning O sits in a small triangle OBC that's itself isosceles, giving OB = OC by the converse fact. That result then feeds a second congruence: in △OAB and △OAC, OB = OC (just shown), ∠OBA = ∠OCA (the same halved angles), and AB = AC (given) — SAS again, so △OAB ≅ △OAC, and CPCT gives ∠OAB = ∠OAC, meaning AO bisects ∠A exactly as required.

What makes this problem satisfying is how it chains two completely separate congruence arguments back to back, each one feeding the next: the first triangle pair establishes OB = OC, and only once that's secured does the second pair become provable by SAS at all. Skipping straight to the second triangle without first establishing OB = OC would leave the proof with only two known facts instead of three, one short of what SAS actually requires. The point O itself is worth a second look, too: it's the meeting point of two angle bisectors dropped from the triangle's two base vertices, and this problem shows that same point automatically lies on the bisector of the third angle as well — all three angle bisectors of any triangle, isosceles or not, are known to meet at a single common point, and this proof is really a specific instance of that broader fact.

A Perpendicular Bisector Forces Two Sides Equal

AD is the perpendicular bisector of BC in △ABC — meaning AD ⊥ BC and D is the midpoint of BC — and the goal is to show AB = AC, i.e., that the triangle is isosceles. In △ADB and △ADC: DB = DC (D is the midpoint), ∠ADB = ∠ADC = 90° (AD is perpendicular to BC), and AD = AD (shared) — SAS again, so △ADB ≅ △ADC, and CPCT gives AB = AC directly.

This problem is worth comparing carefully against the theorem's own proof above, because the two look similar but start from opposite ends: the theorem begins with AB = AC already given and constructs the bisector to prove the angles equal, while this problem begins with the perpendicular bisector already given and proves the sides equal instead. Both routes lean on the exact same SAS pattern — two sides and an included right angle, or two sides and an included bisected angle — which is a strong hint that "isosceles triangle," "angle bisector from the apex," and "perpendicular bisector of the base" are really three descriptions of the same single symmetric figure, each one implying the other two.

Equal Altitudes on Equal Sides

ABC is isosceles with AB = AC, and altitudes BD (to AC) and CE (to AB) are drawn; the task is to show these two altitudes are equal. In △BEC and △CDB: ∠BEC = ∠CDB = 90° (both are altitudes), ∠EBC = ∠DCB (angles opposite the equal sides AC and AB, by the theorem above), and BC = CB — the same segment shared by both triangles. Two angles and a non-included side, AAS, so △BEC ≅ △CDB, and CPCT gives CE = BD — the two altitudes are equal.

The reverse problem sits right alongside this one and is worth stating together with it: if altitudes BD and CE onto AC and AB are instead given equal to begin with, the same AAS pattern applied to △ABD and △ACE — a common angle at A, equal altitudes, and equal right angles — proves AB = AC, meaning the triangle must be isosceles after all. Equal base sides force equal altitudes, and equal altitudes force equal base sides right back — the implication runs both ways, neither direction more fundamental than the other.

Two Isosceles Triangles Sharing One Base

△ABC and △DBC are two separate isosceles triangles built on the same base BC, with AB = AC and DB = DC; the goal is to show ∠ABD = ∠ACD. From △ABC, ∠ABC = ∠ACB (angles opposite equal sides). From △DBC, ∠DBC = ∠DCB, for the same reason. Adding these two equal pairs together, ∠ABC + ∠DBC = ∠ACB + ∠DCB — and the left side is exactly ∠ABD while the right side is exactly ∠ACD, so ∠ABD = ∠ACD follows immediately.

No congruent triangles get named anywhere in this particular proof — it leans entirely on the base theorem applied twice, once to each isosceles triangle, and then a simple addition of angle equalities. That makes it a useful checkpoint: after four problems that each reached for SAS or AAS on a freshly built pair of triangles, this one is a reminder that the isosceles-angle theorem is often powerful enough to finish a proof entirely on its own, without any further congruence argument needed at all.

SSS and RHS Are Still Ahead

Every proof in this exercise reached for SAS or AAS, both of which were already available from the introduction. Exercise 7.3 introduces the two remaining rules, SSS and a formal statement of RHS, completing the full four-rule toolkit that Exercise 7.4 then builds on with inequality relationships between a triangle's sides and angles.