What it does
The mid-point theorem says: in any triangle ABC, if D is the mid-point of AB and E is the mid-point of AC, then DE is parallel to BC and DE = ½ BC. This page lets you test that on as many triangles as you like. Drag the blue corners and the page measures DE, BC, their ratio and the corresponding angles ∠ADE and ∠ABC as you go.
It has three views:
- Theorem: the triangle with D, E and the segment DE, plus a five-step proof (extend DE to F, congruent triangles, a parallelogram).
- Converse: a line through the mid-point D of AB, drawn parallel to BC, meets AC at F. The page measures AF and FC to show F is the mid-point.
- Quadrilateral: join the mid-points P, Q, R, S of the four sides of any quadrilateral. The result is always a parallelogram (Varignon’s theorem), its sides are half the diagonals, its perimeter equals AC + BD and its area is half of ABCD.
It fits the Class 9 mathematics chapter on quadrilaterals (properties of parallelograms and the mid-point theorem). To sort shapes into squares, kites and trapeziums first, use the quadrilateral explorer; the same idea grows into the basic proportionality theorem in similar triangles.
How to use it
- Pick Theorem, Converse or Quadrilateral at the top.
- Drag any blue corner. Corners snap to the grid, and each square is 1 cm, so the lengths are easy to check with Pythagoras.
- Read the measurements under the picture. Lengths are rounded to 2 decimal places.
- In the triangle views, the Right angle, Obtuse and Very thin buttons give quick starting shapes.
A worked example
The page starts with A(6, 2), B(2, 12) and C(15, 11) in centimetres.
- D, the mid-point of AB, is ((6 + 2) ÷ 2, (2 + 12) ÷ 2) = (4, 7). E, the mid-point of AC, is (10.5, 6.5).
- DE = √(6.5² + 0.5²) = √42.5 ≈ 6.52 cm and BC = √(13² + 1²) = √170 ≈ 13.04 cm.
- DE ÷ BC = 0.5 exactly, because √170 = 2 × √42.5.
- ∠ADE and ∠ABC both measure 63.8°, so DE ∥ BC (corresponding angles are equal).
In the Quadrilateral view the starting shape A(3, 3), B(2, 12), C(14, 11), D(12, 2) has diagonals AC ≈ 13.6 cm and BD ≈ 14.14 cm, so the mid-point parallelogram has perimeter about 27.74 cm, and ABCD’s area of 95 cm² gives it an area of 47.5 cm².
Check yourself
Four fresh questions each time: DE from BC, BC from DE, the perimeter of the triangle made by joining all three mid-points, and the perimeter of the mid-point parallelogram of a quadrilateral. Nothing is saved.
FAQ
Does it work for every triangle? Yes: acute, right-angled, obtuse, very thin. That is what makes it a theorem. Dragging only lets you test cases; the proof shows why it can never fail.
What shape do the mid-points of a rectangle make? A rhombus, because a rectangle’s diagonals are equal. A rhombus gives a rectangle, and a square gives a square.
Is this from the NCERT book? It covers the same topic with fibub’s own pictures, proof wording and examples. fibub is not part of NCERT.
Where the numbers come from
- The mid-point theorem, its converse and Varignon's theorem are standard Euclidean geometry. Wording, pictures, proof write-up and examples are fibub's own.
- Lengths, angles and areas checked against Python (math.dist, the shoelace formula) for the example shapes and for 500 random triangles and quadrilaterals.
Standard scientific values, such as constants, atomic weights and melting points, are facts found in every textbook; we name where ours were checked. If you are an author or publisher and want a value, a credit or a table changed or removed, tell us and we will act quickly.
Last updated 2026-11-09.