7 min readConcept sheetPart II (2 marks) and Part III (5 marks) problems

Concurrency theorems: Ceva's theorem and Menelaus' theorem

Ceva: three cevians meet at one point when the side-ratio product is 1. Menelaus: three points on the sides (extended if needed) are collinear when it is −1 (signed lengths).

CevianA segment from a vertex to the opposite side, such as a median, altitude or angle bisector

Ceva's theoremAD, BE, CF are concurrent if and only if (BD/DC)(CE/EA)(AF/FB) = 1

Menelaus' theoremP, Q, R are collinear if and only if (BP/PC)(CQ/QA)(AR/RB) = −1 (directed segments)

UsesMedians meet at the centroid; angle bisectors meet at the incentre

Words you need

A cevian is a line segment from a vertex of a triangle to the opposite side. The end point may also lie on that side extended. Medians, altitudes and angle bisectors are all cevians. Lines are concurrent if they all pass through one point. Points are collinear if they all lie on one straight line.

Ceva vs Menelaus

Ceva's theoremD, E, F lie on BC, CA, AB. The cevians AD, BE and CF are concurrent if and only if (BD/DC) × (CE/EA) × (AF/FB) = 1.

Menelaus' theoremP, Q, R lie on BC, CA, AB or their extensions. P, Q and R are collinear if and only if (BP/PC) × (CQ/QA) × (AR/RB) = −1, using directed (signed) segments. With plain lengths, the product is 1.

Both go round the triangle in the same order: vertex, point, vertex, point. Ceva is about lines meeting; Menelaus is about points on one line. In Menelaus, at least one point lies on a side extended.

Example 1: the medians are concurrent

Problem

Use Ceva's theorem to show that the medians of a triangle meet at one point.

Working

Let AD, BE and CF be the medians of ΔABC. D, E and F are the midpoints of BC, CA and AB.
So BD = DC, CE = EA and AF = FB.
(BD/DC) × (CE/EA) × (AF/FB) = 1 × 1 × 1 = 1
By Ceva's theorem, AD, BE and CF are concurrent.

Answer

The medians are concurrent. They meet at the centroid.

Example 2: find the missing parts (Ceva)

Problem

In ΔABC, the cevians AD, BE and CF are concurrent. BC = 12 cm, AF/FB = 3/4 and CE/EA = 2/3. Find BD and DC.

Working

Ceva: (BD/DC) × (CE/EA) × (AF/FB) = 1
(BD/DC) × (2/3) × (3/4) = 1
(BD/DC) × (1/2) = 1, so BD/DC = 2
BD : DC = 2 : 1 and BD + DC = 12. Total parts = 3.
BD = (2/3) × 12 = 8 cm and DC = 12 − 8 = 4 cm

Answer

BD = 8 cm and DC = 4 cm

Example 3: a straight line across a triangle (Menelaus)

Problem

A straight line cuts AB at R and AC at Q, and meets BC extended beyond C at P. AR/RB = 1/2, CQ/QA = 3/4 and BC = 5 cm. Find PC.

Working

Menelaus with plain lengths: (BP/PC) × (CQ/QA) × (AR/RB) = 1
(BP/PC) × (3/4) × (1/2) = 1, so BP/PC = 8/3
P is beyond C, so BP = BC + CP = 5 + PC.
(5 + PC)/PC = 8/3, so 15 + 3PC = 8PC
5PC = 15, so PC = 3 cm and BP = 8 cm
With signed lengths, BP/PC = −8/3, and the product is −1, as the theorem states.

Answer

PC = 3 cm

In real life

A triangular cardboard sheet balances on a pencil tip at its centroid, where the three medians meet. Ceva's theorem proves that the medians always meet at one point.

Exam angle

2 marks: “Using Ceva's theorem, show that the medians of a triangle are concurrent.” Each median ends at a midpoint, so each ratio is 1. The product is 1 × 1 × 1 = 1. By Ceva's theorem, the medians are concurrent.

2 marks: “In ΔABC, AD, BE and CF are concurrent. BD/DC = 3/2 and CE/EA = 1/2. Find AF/FB.” (3/2) × (1/2) × AF/FB = 1, so (3/4) × AF/FB = 1 and AF/FB = 4/3.

5 marks: “Using Ceva's theorem and the angle bisector theorem, show that the angle bisectors of a triangle are concurrent.” By the angle bisector theorem, BD/DC = AB/AC, CE/EA = BC/BA and AF/FB = CA/CB. The product is (AB × BC × CA)/(AC × BA × CB) = 1. So the bisectors are concurrent, at the incentre.

Remember it

Ceva: lines Concurrent, product 1. Menelaus: points on a line, product −1 with signs. Always go round: vertex, point, vertex, point.