Methods and worked examples: Quadrilaterals

Quadrilaterals24 min readLesson 2 of 7

2.1 Deduction and conjecture A property seen in a few drawings is a conjecture, not a proof. Measuring five rectangles suggests that their diagonals bisect each other, but measurement cannot rule out the ten thousandth rectangle. The textbook builds each property from earlier results with congruence and parallel-line reasoning. The methods on this page follow the same path. The tools are the congruence conditions SAS, ASA, AAS and SSS, the angle sum of a triangle, alternate and corresponding angles on parallel lines, and the linear pair. Each proof below names the two triangles, states why they are congruent, and reads the wanted equality from corresponding parts.

2.2 Proving the rectangle diagonal properties Deduction 1: the diagonals of a rectangle are equal. In rectangle ABCD, consider △ADC and △DAB. AB = CD because opposite sides of a rectangle are equal. AD is common. ∠BAD = ∠CDA = 90°. Therefore △ADC ≅ △DAB by SAS. Corresponding parts give AC = BD, so the diagonals are equal. Deduction 2: the diagonals bisect each other. Let the diagonals meet at O. In △AOB and △COD, ∠OAB = ∠OCD and ∠OBA = ∠ODC because AB is parallel to CD and the diagonals are transversals. AB = CD, so the triangles are congruent by ASA. Hence OA = OC and OB = OD: each diagonal is divided into two equal halves at O.

O A = O C, O B = O D

The diagonals of a rectangle bisect each other

Deduction 3: equal diagonals that bisect each other give a rectangle. Join the endpoints of two equal segments that share a midpoint. The four triangles around the midpoint are congruent in pairs, so the opposite sides of the quadrilateral are equal and the opposite angles are equal. The angle at each vertex is the sum of one acute and one complementary acute angle, which is 90°. So all four angles are right angles and the figure is a rectangle. Deduction 4: all right angles force a rectangle. If every angle of ABCD is 90°, join BD. Triangles BAD and DCB have two right angles and a common side, so they are congruent by AAS. Hence AD = CB and AB = CD, which is the rectangle definition. This shows that "all angles are 90°" is enough to define a rectangle.

2.3 Square from its diagonals Deduction 5: the diagonals of a square are equal and meet at 90°. In square ABCD with diagonals meeting at O, △BOA and △BOC are congruent by SSS: BO is common, OA = OC because diagonals bisect each other, and AB = BC because all sides of a square are equal. Hence ∠BOA = ∠BOC. These two angles form a linear pair and are equal, so each is 90°.

∠ B O A = ∠ B O C = 9 0^{\circ}

The diagonals of a square meet at right angles

Conversely, if the diagonals of a quadrilateral are equal, bisect each other, and meet at 90°, the four small right triangles are congruent. All four sides are equal and all four angles are 90°, so the figure is a square. To construct a square with diagonal 8 cm: draw a segment 8 cm long, locate its midpoint, draw the perpendicular bisector, mark 4 cm on each side of the midpoint, and join the four endpoints in order.

2.4 Rhombus A rhombus is a quadrilateral with all four sides equal. Begin with two equal adjacent sides at a chosen angle, then complete the figure by cutting arcs of the same length from the two free endpoints. The result is a rhombus with that angle at two opposite vertices and its supplement at the other two. Deductions 9 and 10. In rhombus GAME, the diagonal GA makes two isosceles triangles because all sides are equal. Congruence shows that the diagonal bisects the angles it meets. Repeating the argument with both diagonals shows that the diagonals bisect each other at 90° and each diagonal bisects two opposite angles. The rhombus therefore has all the parallelogram properties plus its diagonals as perpendicular angle bisectors.

Worked example 4. A rhombus ABCD has one angle 50°. Find its other angles. Let ∠A = 50°. The diagonal BD is also an angle bisector, so in △ADB the two equal sides AD and AB make the base angles equal. The angle at A is 50°, so each base angle is (180 - 50)/2 = 65°. Since the diagonal bisects the angles at A and C, ∠A = 50° and ∠C = 50°. Opposite angles are equal, so ∠B = ∠D = 180 - 50 = 130°. Check the sum: 50 + 130 + 50 + 130 = 360.

2.5 Kite and trapezium Label a four-sided figure so that the two sides meeting at B are equal (AB = BC) and the two meeting at D are equal (CD = DA); such a figure is called a kite. The diagonal BD joins the two vertices where the equal pairs meet. Triangles AOB and COB are congruent by SSS: BA = BC, AO = OC (the diagonal BD bisects AC), and BO is common. Hence the diagonal BD bisects ∠ABC and ∠ADC, perpendicularly bisects AC, and the two triangles on either side are mirror images. In a trapezium, at least one pair of opposite sides runs parallel. If PQ is parallel to SR, then the angles on each non-parallel side are co-interior and add to 180°: ∠P + ∠S = 180° and ∠Q + ∠R = 180°. In an isosceles trapezium the non-parallel sides are equal. Drawing perpendiculars from the ends of the shorter parallel side to the longer side produces a rectangle and two congruent right triangles, which proves that the angles opposite the equal sides are equal.

2.6 Worked example 5: isosceles trapezium Worked example 5. In isosceles trapezium ABCD, AB is parallel to DC and AD = BC. If ∠D = 100°, find the other angles. Because AB is parallel to DC, ∠A and ∠D are co-interior, so ∠A = 180 - 100 = 80°. The trapezium is isosceles, so the angles at the other end of the equal sides match: ∠B = ∠A = 80°. Finally ∠C = 180 - ∠B = 100°. Check: 80 + 80 + 100 + 100 = 360. The two angles on each parallel side are supplementary, as expected.

2.7 Classifying a quadrilateral To classify a figure, test the families from the strongest conditions to the weakest. A square needs four right angles and four equal sides. A rectangle needs four right angles. A rhombus needs four equal sides. A parallelogram needs two pairs of parallel sides. A kite needs two pairs of adjacent equal sides. A trapezium needs one pair of parallel sides. A figure can pass more than one test. Every square passes the rectangle, rhombus and parallelogram tests, and the Venn diagram records the overlaps. When a question asks whether a statement is always true, look for a counterexample: a rhombus with a 60° angle is a kite but not a rectangle, and a rectangle with unequal sides is a parallelogram but not a rhombus.

in hinglish

Properties ko sirf dekh kar mat maano; deduction se prove karo. Congruence ke conditions SAS, ASA, AAS aur SSS se triangles congruent dikha kar corresponding sides aur angles barabar nikalte hain. Rectangle ke diagonals equal hote hain aur ek doosre ko bisect karte hain; baat ulti bhi sach hai: equal diagonals jo bisect karein, woh rectangle banta hai. Square ke diagonals equal hote hain, bisect karte hain aur 90° par milte hain; har diagonal angle ko 45° + 45° mein baantti hai. Rhombus mein chaaron sides equal hoti hain, diagonals 90° par bisect karti hain aur angles ko bhi bisect karti hain. Rhombus ke ek angle 50° ho to other angles 130°, 50°, 130° aate hain. Kite mein do adjacent equal pairs hote hain; jo diagonal unke beech hai woh doosri diagonal ko perpendicular bisect karti hai aur angles ko bisect karti hai. Trapezium mein parallel sides par co-interior angles 180° ke hote hain; isosceles trapezium mein equal sides ke opposite angles equal hote hain. Classification ke liye strong condition pehle test karo aur counterexample dhoondho.