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Right Triangle Altitude Theorem

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 CC round the semicircle on ABAB, or AA or BB; the readout under the canvas shows CD2CD^2 beside AD⋅DBAD \cdot DB and AC2AC^2 beside AD⋅ABAD \cdot AB. 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?The acute angles α and β add up to 90°, so the altitude CD splits the right angle into β at the A side and α at the B side. The two small triangles ACD and CBD have the angles α, β, 90° like the whole triangle: three similar triangles. From ACD ~ CBD, CD² = AD·DB; from ACD ~ ABC, AC² = AD·AB; from CBD ~ ABC, BC² = BD·AB. Drag C round the semicircle, or A or B.Right triangle altitude
Display
Drag C round the semicircle on AB, or A or B
ABCD
CD² = 7.19, AD·DB = 7.19, AC² = 10.86, AD·AB = 10.86
Step 1 / 6

Step explanations

1A right triangle and the altitude to the hypotenuse
Given
Triangle ABC has a right angle at C; CD is the altitude to the hypotenuse AB, with D between A and B. To prove: CD² = AD · DB (the altitude theorem) and AC² = AD · AB, BC² = BD · AB (the leg theorems).
Drag C round the semicircle: the angle stays right. Learn more about the statement

Getting Started with the Visualizer
The Statement
The Two Acute Angles
The Altitude Theorem
The First Leg
The Second Leg
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 CC round the semicircle to change the shape of the right triangle, or drag AA or BB, and watch each square in the readout stay equal to its product. 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 — C slides round the semicircle on AB, A and B move freely; the hollow foot D follows
4Readout — the line under the figure: CD² beside AD·DB and AC² beside AD·AB, 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 two small triangles — ACD at the A side and CBD at the B side, each with the angles α, β, 90°
2The altitude — CD, the geometric mean of AD and DB; the right angle at D is marked
3The copied angles — β and α at C, the acute angles of the triangle swapped
4Justification tags — Given, Angle sum, AA similarity (three times), Adding
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, drop the altitude CDCD to the hypotenuse ABAB. It cuts the hypotenuse into two pieces ADAD and DBDB, and three relations hold:

CD2=AD⋅DB,AC2=AD⋅AB,BC2=BD⋅ABCD^2 = AD \cdot DB, \qquad AC^2 = AD \cdot AB, \qquad BC^2 = BD \cdot AB


The first is the altitude theorem (Euclid VI.8, corollary): the altitude is the geometric mean of the two pieces. The other two are the leg theorems: each leg is the geometric mean of the hypotenuse and the piece next to that leg. Adding the leg theorems gives the Pythagorean theorem, so this page is also a second proof of it, by similarity instead of by rearrangement.

All three come from one picture: the altitude cuts the right triangle into two smaller right triangles, and all three triangles are similar. In the player CC slides round the semicircle on ABAB, which keeps the angle at CC right by Thales' theorem; DD is the foot of the altitude. The readout shows CD2CD^2 beside AD⋅DBAD \cdot DB and AC2AC^2 beside AD⋅ABAD \cdot AB.
ABCD
Opening scene, frozen

The right triangle with the altitude CD to the hypotenuse; right angles at C and D.


The Two Acute Angles

Call the acute angles of the triangle α\alpha at AA and β\beta at BB. By the angle sum, α+β=90°\alpha + \beta = 90°. Now look at the two small triangles. In ACDACD the angle at DD is right and the angle at AA is α\alpha, so the angle at CC is 90°−α=β90° - \alpha = \beta. In CBDCBD the angle at DD is right and the angle at BB is β\beta, so the angle at CC is 90°−β=α90° - \beta = \alpha:

∠ACD=β,∠DCB=α\angle ACD = \beta, \qquad \angle DCB = \alpha


The altitude splits the right angle at CC into the two acute angles of the triangle, swapped: β\beta on the side of AA, α\alpha on the side of BB. Each small triangle therefore has the angles α\alpha, β\beta, 90°90°, exactly like the whole triangle.
αββαABCD
The two acute angles, frozen

α at A and β at B; the altitude splits the right angle into β and α.


The Altitude Theorem

Compare the two small triangles ACDACD and CBDCBD. Both have the angles α\alpha, β\beta and 90°90°, so by AA similarity they are similar. Match the vertices by their angles: α\alpha sits at AA in the first and at CC in the second, β\beta at CC and at BB, the right angle at DD in both, so A↔CA \leftrightarrow C, C↔BC \leftrightarrow B, D↔DD \leftrightarrow D. Corresponding sides are in proportion:

ADCD=CDBD,CD2=AD⋅DB\frac{AD}{CD} = \frac{CD}{BD}, \qquad CD^2 = AD \cdot DB


The altitude is the geometric mean of the two pieces of the hypotenuse.
αββαABCD
The altitude theorem, frozen

ACD and CBD shaded: ACD ~ CBD, so CD² = AD·DB.


The First Leg

Now compare the small triangle ACDACD with the whole triangle ABCABC. They share the angle α\alpha at AA, and both have a right angle, at DD and at CC. By AA they are similar, with A↔AA \leftrightarrow A, C↔BC \leftrightarrow B, D↔CD \leftrightarrow C:

ADAC=ACAB,AC2=AD⋅AB\frac{AD}{AC} = \frac{AC}{AB}, \qquad AC^2 = AD \cdot AB


The leg ACAC is the geometric mean of the whole hypotenuse and the piece ADAD next to it.
αββαABCD
The first leg, frozen

ACD ~ ABC: AC² = AD·AB.


The Second Leg

In the same way CBDCBD and ABCABC share the angle β\beta at BB and both have a right angle, so they are similar, with C↔AC \leftrightarrow A, B↔BB \leftrightarrow B, D↔CD \leftrightarrow C:

BDBC=BCAB,BC2=BD⋅AB\frac{BD}{BC} = \frac{BC}{AB}, \qquad BC^2 = BD \cdot AB


The leg BCBC is the geometric mean of the hypotenuse and the piece BDBD next to it.
αββαABCD
The second leg, frozen

CBD ~ ABC: BC² = BD·AB.


Why It Holds

Three similar triangles, three proportions, three products. Adding the two leg theorems gives

AC2+BC2=AD⋅AB+BD⋅AB=(AD+DB)⋅AB=AB2AC^2 + BC^2 = AD \cdot AB + BD \cdot AB = (AD + DB) \cdot AB = AB^2


because DD lies between AA and BB. That is the Pythagorean theorem, proved here by similarity; the rearrangement proof on its own page uses areas instead. The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the given figure, the angle sum, AA similarity three times, and an addition.

The whole page rests on one observation, that the altitude copies the two acute angles into the small triangles. Everything else is reading off proportions.
αββαABCD
The conclusion, frozen

Adding the leg theorems: AC² + BC² = AB².


Dragging the Figure

Drag CC round the semicircle: the right angle stays right, the altitude and its foot DD move, the two pieces of the hypotenuse trade length, and in the readout CD2CD^2 stays equal to AD⋅DBAD \cdot DB while AC2AC^2 stays equal to AD⋅ABAD \cdot AB. When CC is at the top of the semicircle the triangle is isosceles, DD is the midpoint and the altitude equals half the hypotenuse. Drag AA or BB to change the hypotenuse; CC keeps its place on the new semicircle.

The player keeps CC above ABAB and the foot DD at least a little way from both ends, so the two small triangles stay visible. A drag that would carry a point out of the frame or squash a small triangle is refused.
αββαABCD
After dragging C, frozen

C moved to the right: AD longer than DB, every square still equal to its product.


Consequences

• Constructing a geometric mean. To draw a segment of length ab\sqrt{ab} with ruler and compass, lay aa and bb end to end, draw the semicircle on the whole, and raise the perpendicular at the join: its height to the semicircle is ab\sqrt{ab}, by the altitude theorem. This is how Euclid squares a rectangle (II.14).
• Pythagoras by similarity. The sum of the two leg theorems is AC2+BC2=AB2AC^2 + BC^2 = AB^2, the proof of the Pythagorean theorem that Euclid gives in Book VI.
• The inverse Pythagorean theorem. Writing the area twice, AC⋅BC=AB⋅CDAC \cdot BC = AB \cdot CD, and combining with the altitude theorem gives 1/CD2=1/AC2+1/BC21/CD^2 = 1/AC^2 + 1/BC^2.
• Trigonometry. With the hypotenuse as the unit, the leg theorems read cos⁡2α=AD\cos^2\alpha = AD and cos⁡2β=BD\cos^2\beta = BD, so cos⁡2α+sin⁡2α=1\cos^2\alpha + \sin^2\alpha = 1 is the sum of the two pieces.
• Thales in reverse. The converse of the altitude theorem holds: if DD on ABAB has CD2=AD⋅DBCD^2 = AD \cdot DB with CD⊥ABCD \perp AB, then the angle at CC is right and CC lies on the circle with diameter ABAB, as on the Thales' theorem page.