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Midpoint of the Hypotenuse

How to use
  1. ▶ Play runs the proof, Next → and ← Back step one scene at a time, Reset restores the opening scene and the original figure. Learn more about the controls
  2. Drag BB along its leg, or AA or CC; the readout under the canvas shows MAMA, MBMB, MCMC and the angle at CC. Learn more about why it holds
  3. Labels toggles the point names, Measures writes each marked angle's size beside its arc. The Step explanations log keeps every step so far. Learn more about reading a scene
Proof?Complete the right triangle ABC to the rectangle CADB by drawing parallels through A and B. A parallelogram with one right angle has four, so CADB is a rectangle; its diagonals AB and CD are equal and bisect each other. So the midpoint M of AB is also the midpoint of CD, and MC = CD/2 = AB/2 = MA = MB. Drag B along its leg, or A or C.Midpoint of the hypotenuse
Display
Drag B along its leg, or A or C (the angle at C stays right)
CABM
MA = 2.03, MB = 2.03, MC = 2.03, ∠C = 90°
Step 1 / 6

Step explanations

1A right triangle and the midpoint of its hypotenuse
Given
Triangle ABC has a right angle at C, and M is the midpoint of the hypotenuse AB, so MA = MB (one tick each). To prove: MC = MA = MB, the midpoint of the hypotenuse is equally far from all three vertices.
Drag B: the angle at C stays right. Learn more about the statement

Getting Started with the Visualizer
The Statement
Completing the Rectangle
Four Right Angles
Equal Diagonals
The Diagonals Bisect Each Other
Why It Holds
Dragging the Figure
Consequences



Getting Started with the Visualizer

DemoA run through the proof player
Step 0 of 7
Step through the proof, then drag BB along its leg, or AA or CC (the angle at CC stays right), and watch the three distances from MM in the readout stay equal. The picture below is the player with its parts numbered; the other tabs list what can be dragged, what each control does and what the step log shows.
Opening scene: the controls
1234567
1Proof — the theorem loaded in the player; the ? gives its idea in one line
2Display — Labels shows the point names, Measures writes each marked angle's size beside its arc
3Draggable points — B slides along the perpendicular to CA at C, A and C move freely; the hollow points D and M follow
4Readout — the line under the figure: MA, MB, MC and the angle at C, live while you drag
5Back, Play, Next, Reset — step through the proof or run it; Reset returns to step 1 and restores the original figure
6Speed and step counter — how fast Play advances, and which of the 6 steps is on screen
7Step explanations — every step reached so far, with its justification tag and a link to the section that explains it; click an entry to jump back to it
Proof complete: step 6 of 6
12345
1The rectangle CADB — shaded, with a right angle at each corner
2The two diagonals — AB and CD, equal and crossing at M
3The circle about M — through A, B and C: the hypotenuse is a diameter
4Justification tags — Given, Construction, Parallelogram angles, Rectangle diagonals, Parallelogram diagonals, Combining
5Replay — at the last step Play becomes Replay and Next is off

The Statement

In a right triangle ABCABC with the right angle at CC, let MM be the midpoint of the hypotenuse ABAB. Then MM is the same distance from all three vertices:

MC=MA=MB=12ABMC = MA = MB = \tfrac12 AB


In other words, the median to the hypotenuse is half the hypotenuse, and MM is the circumcentre of the triangle. This is the converse direction of Thales' theorem: Thales says an angle standing on a diameter is right, this page says the hypotenuse of a right triangle is a diameter of its circumcircle.

The proof completes the triangle to a rectangle and reads the answer off its diagonals, using the parallelogram angles, rectangle diagonals and parallelogram diagonals pages. In the player CC and AA are free and BB slides along the perpendicular to CACA at CC, so the angle at CC stays right without assuming any circle; the readout shows the three distances from MM.
CABM
Opening scene, frozen

The right triangle ABC and the midpoint M of the hypotenuse, MA = MB.


Completing the Rectangle

Through AA draw the line parallel to CBCB, and through BB the line parallel to CACA; they meet at a point DD. The quadrilateral CADBCADB has both pairs of opposite sides parallel (the chevrons), so it is a parallelogram. The triangle is half of it, cut off by the diagonal ABAB.
CABDM
Completing the rectangle, frozen

Parallels through A and B meet at D: CADB is a parallelogram.


Four Right Angles

Neighbouring angles of a parallelogram add up to 180°180° (parallelogram angles). The angle at CC is 90°90°, so its neighbours at AA and BB are 180°−90°=90°180° - 90° = 90°, and the angle at DD, opposite CC, equals it. All four angles are right:

CADB is a rectangleCADB \text{ is a rectangle}
CABDM
Four right angles, frozen

One right angle makes all four right: a rectangle.


Equal Diagonals

Draw the second diagonal CDCD. The diagonals of a rectangle are equal (rectangle diagonals):

CD=ABCD = AB
CABDM
Equal diagonals, frozen

The second diagonal CD, equal to AB.


The Diagonals Bisect Each Other

A rectangle is a parallelogram, and the diagonals of a parallelogram bisect each other (parallelogram diagonals): they cross at a point that is the midpoint of both. The midpoint of ABAB is MM, so the crossing point is MM, the diagonal CDCD passes through MM, and

MC=MD=12CDMC = MD = \tfrac12 CD
CABDM
The diagonals bisect each other, frozen

CD passes through M: MC = MD.


Why It Holds

Put the pieces together:

MC=12CD=12AB=MA=MBMC = \tfrac12 CD = \tfrac12 AB = MA = MB


MM is equally far from AA, BB and CC, so the circle about MM through AA passes through BB and CC: the hypotenuse is a diameter of the circumcircle. The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the given figure, a construction, parallelogram angles, rectangle diagonals, parallelogram diagonals and combining.

The rectangle is the whole idea: a right triangle is half a rectangle, and the centre of a rectangle is equally far from its four corners.
CABDM
The conclusion, frozen

MC = MA = MB: the circle about M passes through all three vertices.


Dragging the Figure

Drag BB along its leg: the triangle grows taller or flatter, the rectangle and its centre MM follow, and in the readout MAMA, MBMB and MCMC stay equal while the angle at CC stays 90°90°. Drag AA or CC to stretch or move the triangle; BB keeps its place along the perpendicular at CC.

The player keeps the legs long enough for the marks and keeps the rectangle and the circle about MM inside the frame; a drag that would push them out is refused.
CABDM
After dragging B, frozen

B slid down its leg: a flatter triangle, M still equally far from all three vertices.


Consequences

• The circumcentre of a right triangle is the midpoint of the hypotenuse, and the circumcentre lies on a side exactly when the triangle is right-angled.
• Thales both ways. Together with Thales' theorem: the angle at CC is right if and only if CC lies on the circle with diameter ABAB.
• The median is half the hypotenuse. In a right triangle the median from the right angle has length c/2c/2; conversely, a triangle in which one median is half the side it meets is right-angled at the vertex it comes from.
• Two isosceles triangles. MC=MAMC = MA and MC=MBMC = MB split the right triangle into two isosceles triangles, so the median from CC makes with the legs the same angles as the hypotenuse does at AA and BB.
• Nine-point circle. The feet of the altitudes see the sides of a triangle at right angles, so by this theorem they lie on circles about the side midpoints; that is the key step of the nine-point circle page.