Chapter 8.5 — Exercise 8.4 — Mid Point Theorem

The Theorem and Its Reverse

In △ABC, if D and E are the midpoints of AB and AC, then DE ∥ BC and DE = ½BC.

The converse runs the logic backward and is just as useful: if a line through the midpoint of one side is drawn parallel to a second side, it automatically bisects the third side too — so in the same triangle, if D is the midpoint of AB and DE ∥ BC, then E must be the midpoint of AC, with no separate measurement needed to confirm it.

Applying the Theorem Twice in a Row

In △ABC, D is on AB with AD = ¼AB, and E is on AC with AE = ¼AC; given DE = 2 cm, the goal is to find BC. Let M and N be the actual midpoints of AB and AC. Since AD = ¼AB = ½(½AB) = ½AM, D is the midpoint of AM — and by the identical reasoning, E is the midpoint of AN. So in △AMN, DE joins two midpoints, giving DE = ½MN, meaning MN = 2 × 2 = 4 cm. Then in △ABC itself, M and N are the midpoints of AB and AC, so MN = ½BC, giving BC = 2 × 4 = 8 cm.

The theorem is applied twice here, once inside a smaller triangle (AMN) nested entirely within the original, and once again in the original triangle itself — each application doubling the previous segment. Recognising that AD = ¼AB is really "D is the midpoint of the midpoint" is the key step that unlocks the whole chain; without spotting that nested structure, the quarter-length condition looks like it has nothing to do with the midpoint theorem at all. The same trick generalizes further still: a point placed at any fraction 1/2ⁿ of a side from the vertex sits at the midpoint of a midpoint of a midpoint, repeated n times, so the theorem could in principle be chained n times over to relate that point back to the original side BC.

Any Quadrilateral's Midpoints Form a Parallelogram

Quadrilateral ABCD has E, F, G, H as the midpoints of AB, BC, CD, DA; the goal is to show EFGH is a parallelogram — a result that holds for absolutely any quadrilateral, convex or not, with no special conditions on ABCD at all. Drawing diagonal AC splits the figure into △ABC and △ADC. In △ABC, E and F are midpoints of AB and BC, so EF ∥ AC and EF = ½AC. In △ADC, H and G are midpoints of AD and CD, so HG ∥ AC and HG = ½AC. Since both EF and HG are parallel to the same line AC and equal in length to each other, EF ∥ HG and EF = HG — one pair of opposite sides of EFGH both equal and parallel, which is enough on its own to make EFGH a parallelogram.

This result is sometimes called the Varignon parallelogram, and what makes it striking is how little it demands of the original figure: ABCD can be any four-sided shape at all, even one with no parallel or equal sides anywhere, and the quadrilateral formed by its four midpoints is still guaranteed to be a well-behaved parallelogram every time.

A Rhombus's Midpoints Form a Rectangle

Starting from the same construction but specifically on rhombus ABCD, with diagonals AC and BD crossing at O: the argument above already gives EFGH as a parallelogram, with EF ∥ AC. Since ABCD is a rhombus, its diagonals AC and BD meet at right angles, so ∠AOB = 90°; and because EH (built the same way from the triangle on the other side) runs parallel to diagonal BD, the angle between EF and EH inherits that same 90° from the angle between AC and BD. A parallelogram with one right angle is a rectangle — so the midpoint quadrilateral of a rhombus is always a rectangle.

Comparing this with the general quadrilateral result above shows exactly what the rhombus's extra structure buys: an arbitrary quadrilateral's midpoint figure is guaranteed to be a parallelogram, full stop, but a rhombus's perpendicular diagonals upgrade that guarantee specifically to a rectangle — the same construction, refined by exactly the one extra fact the rhombus supplies.

Joining Opposite Midpoints Instead of Adjacent Ones

A related question asks about the two segments joining midpoints of opposite sides of a quadrilateral — E to G, and F to H, rather than the EFGH boundary itself — and whether they bisect each other. Since EFGH has already been shown to be a parallelogram for any quadrilateral ABCD, and a parallelogram's own diagonals always bisect each other, EG and FH — which are exactly the diagonals of parallelogram EFGH — must bisect each other too. No new construction is needed at all; the result falls directly out of the Varignon parallelogram already established, treating EFGH's own diagonals rather than its sides.

A Right Triangle's Hypotenuse Midpoint

Right triangle ABC has its right angle at C, and M is the midpoint of the hypotenuse AB; a line through M parallel to BC meets AC at D. Three things need showing: D is the midpoint of AC, MD ⊥ AC, and CM = MA = ½AB. Since M is the midpoint of AB and MD ∥ BC, the converse of the midpoint theorem gives that D must be the midpoint of AC directly — the first result, with no further work. Since MD ∥ BC and AC is a transversal crossing both, ∠ADM = ∠ACB = 90° (corresponding angles), so MD ⊥ AC — the second result.

For the third: in △ADM and △CDM, AD = CD (D is now known to be the midpoint of AC), ∠ADM = ∠CDM = 90° (just shown), and DM = DM (shared) — SAS, so △ADM ≅ △CDM, and CPCT gives AM = CM. Since M is the midpoint of AB, AM already equals ½AB, so CM = AM = ½AB follows immediately. This closing problem quietly ties the whole exercise back to Chapter 7: once D was located using the midpoint theorem alone, everything else needed to finish the proof came from an ordinary SAS congruence, the same rule used constantly since the very first exercise on triangles.

What the Midpoint Quadrilateral Becomes

Leaving Shapes Behind for Numbers

Triangles and quadrilaterals have supplied every construction across these two chapters — congruent halves split by a diagonal, midpoints joined into smaller shapes, angle sums fixed by the number of sides. Chapter 9, Statistics moves away from geometric figures entirely, turning instead to organizing and summarizing collected data — a genuinely different kind of mathematics, though it still leans on the same careful, step-by-step reasoning built up across every proof in this chapter.