Chapter 8.3 — Exercise 8.2 — Parallelograms

One Diagonal, Two Congruent Halves

In parallelogram ABCD, draw diagonal BD. In △ABD and △CDB: AB ∥ DC and BD is a transversal, so ∠ABD = ∠CDB (alternate angles); AD ∥ BC and BD is again a transversal, so ∠ADB = ∠CBD (alternate angles); and BD = DB, the same segment shared by both triangles. Two angles and the included side, ASA, so △ABD ≅ △CDB.

CPCT then delivers both stated properties at once from this single congruence: AB = CD and AD = BC (opposite sides equal), and ∠A = ∠C follows too, since ∠A is made up of ∠ABD... rather, since ∠DAB corresponds directly to ∠BCD across the congruent pair. A second application of the identical argument, drawing the other diagonal AC instead of BD, would establish ∠B = ∠D by the same reasoning — so a single diagonal, chosen either way, is genuinely enough to unlock every one of the four "opposite parts equal" facts this chapter's introduction had simply stated without proof. Notice that neither diagonal needs to be drawn as a special or unusual line — any diagonal of any parallelogram automatically creates this alternate-angle pairing, purely because opposite sides are already parallel by definition, so the two triangles it produces are guaranteed congruent every single time, not just for a conveniently drawn example figure.

A Rectangle Bolted Onto a Parallelogram

ABCD is a parallelogram and ABEF is a rectangle sharing the side AB; the goal is to show △AFD ≅ △BEC. In △AFD and △BEC: ∠AFD = ∠BEC = 90° (every angle of rectangle ABEF is a right angle), AD = BC (opposite sides of parallelogram ABCD), and AF = BE (opposite sides of rectangle ABEF) — hypotenuse and one side matching in two right triangles, RHS, so △AFD ≅ △BEC.

What makes this problem work is that two entirely different quadrilaterals, a parallelogram and a rectangle, are contributing one matching fact each to the same pair of triangles — AD = BC comes purely from the parallelogram's own properties, AF = BE comes purely from the rectangle's, and the shared right angle comes from the rectangle again. Neither shape alone hands over enough facts for RHS; only reading both figures together completes the three conditions the rule needs. This kind of composite figure — two named quadrilaterals sharing a single side, each contributing its own separate set of properties — comes up often enough in later geometry that it's worth treating as its own small skill: identify which shape each fact traces back to before trying to assemble them into a single congruence argument, rather than searching the whole figure for three matching facts without first sorting out where each one actually comes from.

A Rhombus Splits Into Four, Not Just Two

Rhombus ABCD has diagonals AC and BD crossing at O; the goal is to show all four resulting triangles are congruent to one another. In △AOD and △AOB: OD = OB (diagonals of a parallelogram — and every rhombus is one — bisect each other), AD = AB (all four sides of a rhombus are equal), and OA = OA (shared) — SSS, so △AOD ≅ △AOB. The identical SSS pattern, cycling one step around the figure each time, gives △AOB ≅ △COB (using OA = OC, AB = CB, OB = OB) and then △COB ≅ △COD (using OB = OD, BC = DC, OC = OC). Chaining all three congruences together: △AOD ≅ △AOB ≅ △COB ≅ △COD — all four triangles formed by a rhombus's diagonals are congruent.

Every one of the three SSS applications above leans on exactly the same two rhombus facts — equal sides and bisecting diagonals — just matched up against a different pair of triangles each time; nothing new about the rhombus needs to be discovered partway through, only the same two facts reapplied three times in sequence around the figure. The four congruent triangles this produces aren't just a curiosity, either — they're the direct reason a rhombus's diagonals end up perpendicular to each other, a fact stated but not proved back in the introduction: since △AOB ≅ △AOD by the chain above, CPCT gives ∠AOB = ∠AOD, and since these two angles sit side by side along the straight line BD, they must also sum to 180°; two equal angles summing to 180° forces each one to be exactly 90°. Every proof across this exercise reaches for one of the four congruence rules established in the previous chapter — mostly ASA and SSS, with RHS making a single appearance in the rectangle problem above — which makes this exercise as much a continuation of Chapter 7's toolkit as it is a genuinely new topic in its own right.

Bisected Angles at Two Vertices Meeting Inside

In quadrilateral ABCD (not assumed to be a parallelogram), the bisectors of ∠C and ∠D meet at point O; the goal is to prove ∠COD = ½(∠A + ∠B). In △COD, ∠OCD = ½∠C and ∠ODC = ½∠D (both bisected angles), and the triangle's own angles sum to 180°, so ∠COD = 180° − (½∠C + ½∠D). Separately, the full quadrilateral's angle sum gives ∠A + ∠B + ∠C + ∠D = 360°, so ∠A + ∠B = 360° − (∠C + ∠D); halving both sides, ½(∠A + ∠B) = 180° − ½(∠C + ∠D). The right-hand sides of both derived equations are now identical, so ∠COD = ½(∠A + ∠B) follows directly.

This last problem is a useful reminder that not every result in this exercise actually needs the parallelogram's special structure at all — it holds for any quadrilateral whatsoever, since the only two facts used throughout are the general 360° angle sum from the introduction and ordinary angle-bisector arithmetic, neither of which depends on any sides being parallel.

It's worth noticing what would change if ABCD were specifically a parallelogram rather than an arbitrary quadrilateral: ∠A = ∠C and ∠B = ∠D would already be known from the theorem proved earlier in this exercise, and since ∠A + ∠B + ∠C + ∠D still totals 360°, the pair ∠A + ∠B would automatically equal 180° — collapsing the formula ∠COD = ½(∠A + ∠B) down to the much simpler ∠COD = 90° in that special case. The general quadrilateral result proved here is genuinely the harder, more useful fact; the parallelogram case is just one easy consequence sitting inside it.

Where the Diagonal Itself Becomes the Subject

This exercise used a diagonal purely as scaffolding to prove facts about a parallelogram's sides and angles. Exercise 8.3 turns the diagonals into the actual subject of the theorem — proving that they bisect each other, and that bisecting diagonals are themselves enough to guarantee a parallelogram in the first place.