10 min readPathway sheetPart IV construction (8 marks), either-or choice

Construction: triangle given base, vertical angle and median, altitude or bisector

Draw the base, then a circle whose arc holds the vertical angle. The median, altitude or bisector then fixes the third vertex on that arc.

Vertical angleThe angle at the vertex opposite the base (∠A when the base is BC)

Key ideaAlternate segment theorem: every point on the upper arc sees BC at the given angle

Common startBase, angle at B, perpendicular, perpendicular bisector, circle

InstrumentsRuler, compasses, protractor, sharp pencil

Steps of construction: base, vertical angle and median

Question: construct ΔABC with BC = 6 cm, ∠A = 60° and median AM = 5 cm. Instruments: ruler, compasses, protractor, sharp pencil.

1
Draw a rough figure. Show ΔABC on base BC, the circle through B and C, and the median AM. Write BC = 6 cm, ∠A = 60° and AM = 5 cm.
2
Draw a line segment BC = 6 cm.
3
At B, draw a ray BX below BC so that ∠CBX = 60°.
4
At B, draw BY ⊥ BX. BY goes above BC.
5
Draw the perpendicular bisector of BC. It cuts BY at O and BC at M, the midpoint.
6
With O as centre and OB as radius, draw a circle. It passes through C. Every point on the arc above BC makes an angle of 60° with B and C.
7
With M as centre and radius 5 cm, draw an arc. It cuts the upper arc at A and A′.
8
Join AB and AC. ΔABC is the required triangle.
ΔA′BC is a mirror image of the same size

Steps of construction: base, vertical angle and altitude

Question: construct ΔABC with BC = 5 cm, ∠A = 50° and the altitude from A to BC = 4.5 cm. Instruments: ruler, compasses, protractor, sharp pencil.

1
Draw a rough figure. Show ΔABC, the circle through B and C, and the altitude AD ⊥ BC. Write BC = 5 cm, ∠A = 50° and AD = 4.5 cm.
2
Draw BC = 5 cm. At B, draw a ray BX below BC with ∠CBX = 50°.
3
At B, draw BY ⊥ BX.
4
Draw the perpendicular bisector of BC. It cuts BY at O and BC at M.
5
With O as centre and OB as radius, draw the circle.
6
On the perpendicular bisector, mark H above BC so that MH = 4.5 cm.
7
Through H, draw a line parallel to BC. It cuts the circle at A and A′.
8
Join AB and AC. ΔABC is the required triangle. Draw AD ⊥ BC to show the altitude of 4.5 cm.

Steps of construction: base, vertical angle and bisector

Question: construct ΔABC with BC = 6 cm and ∠A = 50°. The bisector of ∠A meets BC at D, with CD = 2.5 cm. Instruments: ruler, compasses, protractor, sharp pencil.

1
Draw a rough figure. Show ΔABC, the circle through B and C, the bisector AD and the lowest point N of the circle. Write BC = 6 cm, ∠A = 50° and CD = 2.5 cm.
2
Draw BC = 6 cm. At B, draw a ray BX below BC with ∠CBX = 50°.
3
At B, draw BY ⊥ BX.
4
Draw the perpendicular bisector of BC. It cuts BY at O.
5
With O as centre and OB as radius, draw the circle. The perpendicular bisector meets the circle below BC at N.
6
Mark D on BC so that CD = 2.5 cm.
7
Join ND and extend it to meet the upper arc at A.
8
Join AB and AC. ΔABC is the required triangle, and AD bisects ∠A.

Why it works

O lies on BY, so OB ⊥ BX. This makes BX a tangent to the circle at B. By the alternate segment theorem, every angle in the upper segment equals ∠CBX, the given angle. So A can be any point on the upper arc, and the median, altitude or bisector picks the right one. N is the middle of the lower arc BC, so arcs BN and NC are equal. Equal arcs give ∠BAN = ∠NAC, so AN bisects ∠A. If the median arc or the parallel line misses the circle, no such triangle exists.

Example: check by measuring

Problem

You have constructed ΔABC with BC = 6 cm, ∠A = 60° and median AM = 5 cm. Find the length of the altitude from A to BC.

Working

From A, draw AD ⊥ BC. With A as centre, cut BC at two points. Then bisect the part between them.
Measure AD with a ruler.
Also check that ∠BAC reads 60° on the protractor and that AM is 5 cm.

Answer

Altitude AD ≈ 4.6 cm (a reading from 4.5 cm to 4.7 cm shows a neat drawing).

In real life

Every seat on a curved arc around a stage sees the stage at the same angle. A photographer standing anywhere on that arc fits the same width of a temple tower into the frame.

Exam angle

8 marks: “Construct ΔPQR in which QR = 5.5 cm, ∠P = 60° and the median PM from P to QR = 4.5 cm. Find the length of the altitude from P to QR.” Rough figure; QR = 5.5 cm; ray QX below QR with ∠RQX = 60°; QY ⊥ QX. The perpendicular bisector of QR meets QY at O. Draw the circle with centre O through Q. An arc of 4.5 cm from M gives P. Altitude measures about 4 cm.

8 marks: “Construct ΔABC with BC = 5.5 cm, ∠A = 45° and the altitude from A to BC = 4 cm.” Steps 1 to 5 as usual, with a 45° angle at B. Mark H on the perpendicular bisector with MH = 4 cm. A line through H parallel to BC cuts the circle at A. Join AB and AC.

8 marks: “Draw ΔABC with BC = 6.5 cm and ∠A = 70°. The bisector of ∠A meets BC at D, with BD = 4 cm.” Steps 1 to 5 as usual, with a 70° angle at B. The perpendicular bisector meets the circle below BC at N. Mark D with BD = 4 cm. Join ND and extend it to the upper arc at A. Join AB and AC.

Remember it

Base, angle below, perpendicular up, bisector across, circle round. Then: median = arc from M; altitude = line parallel to the base; bisector = line from N through D.