Midpoint Theorem

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The Midpoint Theorem

It is pronounced by the way of the Midpoint Theorem that the line segment connecting the midpoints of any two sides of a triangle is parallel to the 3rd side and equal to half of the 3rd side. For example, let’s take an arbitrary triangle, ΔABC. Suppose that D and E are the midpoints of line segment AB and AC. Now assuming that you join D and E:

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The statement of Midpoint Theorem declares that DE will be parallel to BC and also equivalent to exactly half of BC.

Midpoint Theorem Proof

To state and prove midpoint theorem, we need to do the following:

According to the Mid-Point Theorem, a segment connecting the mid-points of two sides of a triangle is parallel to the 3rd side and equal to half the 3rd side.

Given: In ΔABC, P and Q are mid-points of line segment AB and AC respectively

To Prove:

i)             PQ || BC

ii)            ii) PQ = ½ BC

Construction: Draw BA || CR to meet PQ generated at R.

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∠QAP = ∠QCR. (Pair of alternate angles) ------ (1)

AQ = QC. (Because, Q is the midpoint of side AC) ----- (2)

∠AQP = ∠CQR (Vertically opposite angles) ----- (3)

Therefore, ΔAPQ ≅ ΔCRQ (ASA Congruence rule)

PQ = QR. (by law of Corresponding parts of Congruent triangles [CPCT]).

 or PQ = 1/ 2 PR ------- (4)

⇒ AP = CR (by CPCT)....(5)

But, AP = BP. (because, P is the midpoint of AB)

⇒ BP = CR

Moreover, BP || CR. (through construction)

In quadrilateral BCRP, BP || CR and BP = CR

Hence, quadrilateral BCRP is a parallelogram.

BC || PQ or BC || PR,

In addition, PR = BC (because, BCRP is a parallelogram)

⇒ 1 /2 PR = 1/ 2 BC

⇒ PQ = 1/ 2 BC. [from (4)]

Now, with the triangle midpoint theorem, take a ΔABC, E and F are the mid-points of side AB and AC respectively.

Construction: - By point C, construct a line II BA to join EF generated at D.

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1. AF=CF (F is midpoint of side AC)

2. ∠AFE= ∠CFD (Vertically opposite angles)

3. ∠EAF= ∠DCF [Alternate. angles, BA II CD(through construction), AC is a transversal]

4. Thus, ΔAEF = CDF (by rule of ASA)

5. EF = FD and AE = CD (by CPCT)

6. AE = BE (E is midpoint of side AB)

7. BE = CD (from equation 5 and 6)

8. EBCD is a IIgm [BA II CD (through construction) and BE = CD (from 7)]

9. EF II BC and ED = BC (Because EBCD is a IIgm)

10. EF = 1/2 ED (Because, EF = FD (from 5)

11. EF = 1/2 BC (Because, ED = BC (from 9)

Therefore, EF II BC AND EF = ½ BC which proves the mid-point theorem.

Solved Examples on The Midpoint Theorem

Example 1: Take a ΔABC, and suppose that D be any point on the side BC. Assume that X and Y be the midpoints of sides AB and AC respectively. Prove that XY will intersect AD.

Solution: By the statement of midpoint theorem, XY || BC. Now, consider triangle ABD. The line segment XE is parallel to the base of the side BD, and X is the midpoint of side AB. Through the converse of the midpoint theorem statement, E should be the midpoint of AD.

Hence, XY intersects AD.

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Example 2: In the adjoining figure, all measurements are reflected in centimetres. Find the length of the side AO if AX measures 10 cm.

Solution: Given in the figure AD = DB = 4cms

AE = EC = 6cms

Then, D and E are the midpoints of AB and AC respectively.

Thus, DE intersects AX at point O

Therefore, AO = ½ AX

= 1/2 × 10 = 5

Thus, the length of AO = 5.

FAQs (Frequently Asked Questions)

1. What Do We Understand By the Converse of the Midpoint Theorem?

Answer: In Euclidean geometry, the converse of the midpoint theorem is as crucial as the theorem itself. The converse of midpoint theorem pronounces that: "If a line segment is drawn crossing across the midpoint of any one side of a triangle and parallel to another side, then this line segment intersects the remaining 3rd side.

2. What is Meant by the Midpoint Theorem?

Answer: The midpoint is the centermost point of a line segment, which also means that it is at the same distance from both the end point.

3. What is the Use of the Midpoint Theorem Formula?

Answer: the midpoint formula is used to identify the coordinates of the midpoint of the line segment. This formula used for midpoint theorem is familiar to high school students.

The point that is equidistant from two points A (x1, y1) and B (x2, y2) on a line segment is known as the midpoint.