Chapter 8.2 — Exercise 8.1 — Quadrilateral Properties

Six Statements, One Family Tree

• Every parallelogram is a trapezium. True — a parallelogram's two pairs of parallel sides already include at least one pair, satisfying the trapezium's weaker requirement.
• All parallelograms are quadrilaterals. True — a parallelogram is a quadrilateral with an extra condition added, never a different category of shape.
• All trapeziums are parallelograms. False — a trapezium only needs one pair of sides parallel; nothing forces the second pair to match.
• A square is a rhombus. True — a square's four equal sides already satisfy everything a rhombus requires.
• Every rhombus is a square. False — a rhombus only fixes the four sides equal; its angles are free to be anything except a right angle, which is the one extra condition a square adds.
• All parallelograms are rectangles. False — a parallelogram needs one 90° angle to become a rectangle, and most parallelograms never have one.

Every "False" answer above fails for the identical structural reason: it tries to run the family-tree chain backward, claiming that a broader category (trapezium, rhombus, parallelogram) automatically satisfies a narrower one (parallelogram, square, rectangle) further down the chain. The chain only ever guarantees inheritance in the forward direction — down toward more specific shapes — never back up toward more general ones. A useful way to picture the whole set is as nested regions on a page rather than as a straight line: parallelograms form one region inside the space of all quadrilaterals, rectangles and rhombuses are two smaller, separate regions inside that parallelogram region, and squares sit in the tiny overlap where the rectangle and rhombus regions meet each other — inside both at once, and therefore inside every larger region surrounding them too.

Every Property Against Every Shape, at Once

Reading down the Square column tells the whole story of this table at a glance: every single row is YES, since a square inherits every property that any of the other four shapes can claim. Reading down the Trapezium column tells the opposite story — only its own defining row is YES, because a general trapezium carries no other guarantee beyond the one parallel pair it's built from. The Rhombus and Rectangle columns are worth comparing side by side too: they never share a YES on the same row except for the properties both inherit from being parallelograms in the first place (rows two through six) — a rhombus earns its own YES rows through equal sides and perpendicular diagonals, a rectangle earns its own through equal diagonals and right angles, and neither shape's specialization helps with the other's. This is really the same overlapping-regions picture from above, now visible one row at a time: a shape that happens to be both a rhombus and a rectangle simultaneously has nowhere else to go except becoming a square, since it would need to satisfy every YES in both columns at once — which is exactly the definition already given for a square.

Equal Legs Force Equal Base Angles in a Trapezium

Trapezium ABCD has AB ∥ CD and AD = BC — a trapezium with its two non-parallel sides (its "legs") equal, sometimes called isosceles. The goal is to show ∠A = ∠B and ∠C = ∠D. Draw perpendiculars CE and DF from C and D onto AB. In △AFD and △BEC: ∠AFD = ∠BEC = 90° (construction), AD = BC (given), and DF = CE (both are the perpendicular distance between the same pair of parallel lines AB and CD, so they must be equal) — hypotenuse and one side, RHS, so △AFD ≅ △BEC, and CPCT gives ∠A = ∠B.

The second half follows from the quadrilateral's own angle relationships rather than from any further congruence: since AB ∥ CD, angles adjacent to the same leg are supplementary, so ∠A + ∠D = ∠B + ∠C. Substituting ∠A = ∠B (just proved) into this equation and cancelling gives ∠D = ∠C directly. This problem is a genuinely useful bridge between the trapezium's own single defining property and the four-rule congruence toolkit from the previous chapter — nothing about a general trapezium guarantees RHS is available, so the equal-legs condition given here is doing real, necessary work, not just decorating the setup.

Reading a Ratio Straight Off the 360° Total

A quadrilateral's four angles are in the ratio 1 : 2 : 3 : 4. Since the ratio's terms add to 1 + 2 + 3 + 4 = 10, and the four actual angles must sum to 360°, each "part" of the ratio is worth 360° ÷ 10 = 36°. The four angles are then 1 × 36° = 36°, 2 × 36° = 72°, 3 × 36° = 108°, and 4 × 36° = 144° — and checking the total, 36 + 72 + 108 + 144 = 360°, confirms the split is correct.

This same one-line technique — divide the fixed total by the sum of the ratio's parts, then scale each part up by that same factor — works for any quadrilateral angle ratio at all, not just this specific 1:2:3:4 split, precisely because the angle sum is fixed at 360° for every convex quadrilateral regardless of its individual shape. The same method applies just as directly to a triangle's fixed 180° total or to any other polygon whose angle sum is already known — only the fixed total itself ever changes between shapes.

What a Rectangle's Diagonal Cuts It Into

Rectangle ABCD has AC as one of its diagonals; the question asks what kind of triangle ACD turns out to be. Since every angle of a rectangle is 90°, ∠D = 90° already, which makes △ACD a right triangle on its own without needing anything further from the diagonal itself. Because ∠CAD and ∠ACD are the triangle's other two angles, they must sum to 90° to keep the triangle's total at 180° — meaning that unless the rectangle happens to be a square, those two remaining angles split unevenly, one larger than the other, rather than both landing at a tidy 45°.

From Definitions to Full Proofs

Every result checked in this exercise so far has come from combining the introduction's stated properties with ordinary angle arithmetic. Exercise 8.2 turns two of those parallelogram properties — equal opposite sides and equal opposite angles — into formally proved theorems for the first time, each one built directly on a diagonal splitting the parallelogram into a pair of congruent triangles.