Construction and proof practice: Quadrilaterals

This page adds practice, not new theory. Every task below is original and uses only the properties proved in the chapter. Write each construction with ruler and compass, label the figure as you build it, and then explain why the finished shape has the required property. For the proofs, set out the given and the claim first, name the two triangles you compare, and state the congruence condition. Use a sharp pencil, keep the compass opening fixed for each arc, and write the radius next to the arc.

5.1 Task 1: Construct a parallelogram from two sides and an included angle

Construct a parallelogram ABCD with AB = 6 cm, AD = 4 cm and angle DAB = 60 degrees.

Solution. Step 1. Draw AB = 6 cm. Step 2. At A, construct a 60 degree ray with the compass: an equilateral triangle built on any convenient segment gives the 60 degree angle. Mark D on this ray with AD = 4 cm. Step 3. With centre B and radius 4 cm draw an arc. With centre D and radius 6 cm draw a second arc. Let the two arcs meet at C. Step 4. Join BC and DC. Why it is a parallelogram. By construction AB = DC = 6 cm and AD = BC = 4 cm. In triangles ABD and CDB, AB = CD, AD = CB and BD is common, so the triangles are congruent by SSS. Hence angle ABD equals angle CDB, so AB is parallel to DC. The same congruence gives angle ADB = angle CBD, so AD is parallel to BC. ABCD is therefore the required parallelogram.

5.2 Task 2: Construct a rhombus from its diagonals

Construct a rhombus whose diagonals are 7 cm and 5 cm.

Solution. Step 1. Draw AC = 7 cm. Step 2. Construct the perpendicular bisector of AC with the compass, and let it meet AC at O. Step 3. On the bisector mark B and D on opposite sides of O with OB = OD = 2.5 cm. Step 4. Join AB, BC, CD and DA. Why it is a rhombus. Each side is the hypotenuse of a right triangle with legs 3.5 cm and 2.5 cm, so all four sides are equal, each about 4.30 cm. A quadrilateral whose diagonals bisect each other at right angles is a rhombus, so ABCD has the required shape.

5.3 Task 3: Angle chasing in a parallelogram

In parallelogram ABCD, angle A = 3x and angle B = x + 40 degrees. Find all four angles.

Solution. Adjacent angles of a parallelogram lie between parallel sides, so they are co-interior angles and add to 180 degrees. Thus 3x + (x + 40) = 180, so 4x = 140 and x = 35. Hence angle A = 105 degrees and angle B = 75 degrees. Opposite angles are equal, so angle C = 105 degrees and angle D = 75 degrees. Check: 105 + 75 + 105 + 75 = 360 degrees.

5.4 Task 4: Proof that the two diagonals of a parallelogram cut each other in half

Given: ABCD is a parallelogram and its diagonals AC and BD meet at O. To prove: AO = OC and BO = OD.

Proof. AB is parallel to DC, so angle OAB = angle OCD and angle OBA = angle ODC, because they are alternate angles. Also AB = DC, since opposite sides of a parallelogram are equal. Therefore triangle AOB is congruent to triangle COD by ASA. Corresponding parts of congruent triangles are equal, so AO = OC and BO = OD. The diagonals bisect each other at O.

5.5 Task 5: Proof that the diagonals of a rhombus are perpendicular

Given: ABCD is a rhombus and its diagonals AC and BD meet at O. To prove: AC is perpendicular to BD.

Proof. A rhombus is a parallelogram, so its diagonals bisect each other: AO = OC and BO = OD. All four sides are equal, so AB = CB. In triangles AOB and COB, AO = CO, OB is common and AB = CB, so the triangles are congruent by SSS. Hence angle AOB = angle COB. These two angles form a linear pair, so each is 90 degrees, and AC is perpendicular to BD.

5.6 Task 6: Angle chasing in a kite

In kite PQRS, PQ = PS and RQ = RS. Angle P = 100 degrees and angle Q = 70 degrees. Find angles R and S.

Solution. In a kite, the two angles between the unequal pairs of sides are equal, so angle S = angle Q = 70 degrees. The angles of a quadrilateral add to 360 degrees, so angle R = 360 - 100 - 70

• 70 = 120 degrees. Check: 100 + 70 + 120 + 70 = 360 degrees.

5.7 Task 7: Proof for an isosceles trapezium

Given: trapezium ABCD with AB parallel to DC and AD = BC. To prove: angle A = angle B.

Construction: draw DE perpendicular to AB and CF perpendicular to AB, meeting AB at E and F.

Proof. DE = CF, because they are the perpendicular distances between the same pair of parallel lines. Also AD = BC (given) and angle AED = angle BFC = 90 degrees. So triangles ADE and BCF are congruent by RHS. Hence angle DAE = angle CBF, that is, angle A = angle B.