The three medians of ∆ABC pass through a common point, which divides each of the medians in the ratio 2 : 1, with the longer part connecting to the vertex. This point is called the centroid of ∆ABC.
Step-by-step solution
To find: Prove that AR, BQ, CP meet at one point M, and that AM : MR = BM : MQ = CM : MP = 2 : 1
Idea: Handle two medians at a time. To get a 2 : 1 split, cut the longer part in half and show the three pieces are equal, using the Midpoint Theorem twice.
- Let medians CP and BQ meet at M. Let X and Y be the midpoints of BM and CM.½ mark
- Midpoint Theorem in ∆ABC: PQ ‖ BC and PQ = ½BC. In ∆MBC: XY ‖ BC and XY = ½BC. So PQ ‖ XY and PQ = XY.1 mark
- ∠MPQ = ∠MYX and ∠MQP = ∠MXY (alternate angles), with PQ = YX between them, so ∆MPQ ≅ ∆MYX (ASA). Hence MQ = MX and MP = MY.1 mark
- MQ = MX = XB, so BM = 2MQ; MP = MY = YC, so CM = 2MP. Thus BM : MQ = 2 : 1 and CM : MP = 2 : 1.1 mark
- Repeat with medians AR and BQ, meeting at N: the same reasoning gives BN : NQ = 2 : 1 and AN : NR = 2 : 1. But exactly one point divides BQ in the ratio 2 : 1, so N = M. So all three medians pass through M, and M divides each in the ratio 2 : 1 from the vertex.1½ marks
Check: A(0, 6), B(−6, 0), C(6, 0): the centroid is ((0 − 6 + 6)/3, (6 + 0 + 0)/3) = (0, 2). On median AR with R(0, 0): AM = 4, MR = 2, ratio 2 : 1 ✓.
Answer to write in the exam
Let CP, BQ meet at M; X, Y midpoints of BM, CM
PQ ‖ BC, PQ = ½BC (Midpoint Theorem in ∆ABC); XY ‖ BC, XY = ½BC (in ∆MBC) ⇒ PQ ‖ XY, PQ = XY
∆MPQ ≅ ∆MYX (ASA: ∠MPQ = ∠MYX, PQ = YX, ∠MQP = ∠MXY, alternate angles)
⇒ MQ = MX = XB and MP = MY = YC ⇒ BM : MQ = 2 : 1, CM : MP = 2 : 1
Similarly AR, BQ meet at N with BN : NQ = 2 : 1 = AN : NR
Only one point divides BQ in ratio 2 : 1 ⇒ N = M
∴ The medians are concurrent at M, and M divides each in the ratio 2 : 1 from the vertex.
Common mistakes that cost marks
- Proving the 2 : 1 ratio for two medians and then just stating the third passes through M. The uniqueness step (N = M) is needed.
- Taking X and Y as midpoints of BQ and CP instead of BM and CM.
- Writing CM : MP = 1 : 2. The longer part is at the vertex.
How this can come in the exam
G is the centroid of ∆ABC and BE is a median with BG = 6 cm. Find GE and BE.
Show answer
BG : GE = 2 : 1 (1 mark), so GE = 3 cm and BE = 9 cm (1 mark).Assertion (A): The centroid of a triangle always lies inside the triangle.
Reason (R): The centroid lies on each median, two-thirds of the way from the vertex.
- Both A and R are true, and R is the correct explanation of A.
- Both A and R are true, but R is not the correct explanation of A.
- A is true but R is false.
- A is false but R is true.
Show answer
(A) Both A and R are true, and R is the correct explanation of A.
Each median lies inside the triangle, and the centroid is a point of the median between its ends, so it is inside. R explains A.
Try one yourself
In ∆PQR, medians PD and QE meet at G. If PD = 15 cm and QE = 12 cm, find PG and GE.
Show answer
PG = ⅔ × 15 = 10 cm; GE = ⅓ × 12 = 4 cm.
More questions like this
- Give another proof of the Centroid Theorem using a clever construction that we will describe only partly and ask you to complete. Once again let M be the intersection point of the two medians CP and BQ. We want to show that the intersection point of line AM with side BC is the midpoint of BC. For this, extend AM up to a carefully chosen point S such that BSCM becomes a parallelogram. (Hint: To arrange MC ‖ BS, first note that PM and MC are along the same line. Now where should M be on segment AS? Does your placement of S also prove MB ‖ CS?)
- The midpoints of the four sides of a quadrilateral are the vertices of a parallelogram.
- (i) If P, Q, R are the midpoints of sides AB, AC, BC respectively of ∆ABC, show that ∆PQR is congruent to ∆QPA and to two other triangles which you should identify.
(ii) Suppose someone erases ∆ABC, leaving only ∆PQR on the paper. Can you reconstruct ∆ABC from ∆PQR? - In ∆ABC, let M and N be midpoints of AB and AC respectively. Let D be any point on BC. Show that MN bisects AD.
- In a quadrilateral ABCD, suppose AB ‖ DC and AB ≠ CD. Suppose G and H are the midpoints of AC and BD respectively. Prove that GH ‖ AB. (Why did we assume AB ≠ CD?)
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