▶ 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
Drag A, B or C; the altitudes, H and the large triangle follow, and the readout under the canvas keeps the angle between CH and AB at 90°. Learn more about why it holds
Proof?Through each vertex draw the line parallel to the opposite side. The three lines make a larger triangle A′B′C′, built from parallelograms, in which A, B and C are the midpoints of the sides. Each altitude of ABC is perpendicular to a side of the big triangle and passes through its midpoint, so the altitudes are the perpendicular bisectors of A′B′C′, and those meet at one point. Drag any vertex.Orthocentre
Display
Drag A, B or C; the altitudes, H and the big triangle follow
angle between CH and AB: 90.0°
Step 1 / 6
Step explanations
1Two altitudes meet at H
Given AD and BE are altitudes of triangle ABC: AD ⟂ BC and BE ⟂ CA. They meet at H. To prove: the altitude from C passes through H as well.
Step through the proof, then drag the vertices and watch the three altitudes keep meeting at one point inside the triangle. 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
Proof?OrthocentreDisplayLabelsMeasuresDrag A, B or C; the altitudes, H and the big triangle follow
angle between CH and AB: 90.0°
← Back▶ PlayNext →ResetNormal ▾Step 1 / 6
Step explanations
1Two altitudes meet at H
Given AD and BE are altitudes of triangle ABC: AD ⟂ BC and BE ⟂ CA. They meet at H. To prove: the altitude from C passes through H as well.
1234567
1Proof — the theorem loaded in the player; the ? gives its idea in one line
2Display — Labels shows the point names
3Draggable points — A, B and C move freely; the hollow feet D, E, F, the crossing point H and the corners of the large triangle follow
4Readout — the line under the figure: the angle between CH and AB, which stays at 90° 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
Proof?OrthocentreDisplayLabelsMeasuresDrag A, B or C; the altitudes, H and the big triangle follow
angle between CH and AB: 90.0°
← Back↻ ReplayNext →ResetNormal ▾Step 6 / 6
Step explanations
1Two altitudes meet at H
Given AD and BE are altitudes of triangle ABC: AD ⟂ BC and BE ⟂ CA. They meet at H. To prove: the altitude from C passes through H as well.
2Through each vertex, the parallel to the opposite side
Construction Draw the line through A parallel to BC, through B parallel to CA, and through C parallel to AB. They meet in pairs at A′, B′, C′ and form a larger triangle A′B′C′ around ABC.
3A, B, C are the midpoints of the sides of A′B′C′
Parallelogram sides ABA′C has both pairs of opposite sides parallel, so it is a parallelogram and BA′ = AC; likewise AC′BC is a parallelogram and C′B = AC. So C′B = BA′: B is the midpoint of C′A′. In the same way A is the midpoint of B′C′ and C the midpoint of A′B′.
4The altitudes are perpendicular bisectors of A′B′C′
Perpendicular to parallels BE ⟂ CA, and C′A′ ∥ CA, so BE ⟂ C′A′; and BE passes through B, the midpoint of C′A′. So the line BE is the perpendicular bisector of C′A′. In the same way AD is the perpendicular bisector of B′C′.
5Perpendicular bisectors meet at one point
Circumcentre The perpendicular bisectors of a triangle meet at one point, its circumcentre. For A′B′C′ two of them are the lines AD and BE, which meet at H; so the third, the perpendicular bisector of A′B′, passes through H too.
6The third altitude passes through H
Same line The perpendicular bisector of A′B′ is the line through its midpoint C perpendicular to A′B′, and A′B′ ∥ AB, so it is the line through C perpendicular to AB: the altitude CF. It passes through H, so all three altitudes meet at H, the orthocentre. ∎
12345
1The orthocentre H — where the three altitudes meet; the circumcentre of the large triangle
2The large triangle — A′B′C′, dashed, with A, B, C as the midpoints of its sides
3The third altitude CF — added at the last step: the perpendicular bisector of A′B′, through H
4Justification tags — Given, Construction, Parallelogram sides, Perpendicular to parallels, Circumcentre, Same line
5Replay — at the last step Play becomes Replay and Next is off
The Statement
An altitude of a triangle is the perpendicular from a vertex to the line of the opposite side. There are three, and the theorem says they pass through one point H, the orthocentre. Unlike the centroid, the circumcentre and the incentre, the orthocentre has no distance property of its own that makes the third line obviously pass through it; the proof on this page, due to Gauss, gets it by borrowing the circumcentre's.
The idea is to draw a second, larger triangle around the first, in such a way that the altitudes of the small triangle become the perpendicular bisectors of the large one. Perpendicular bisectors are known to be concurrent, and that does it. In the player the feet of the altitudes, the crossing point H and the corners of the large triangle are hollow, computed from A, B, C. The readout under the canvas shows the angle between CH and AB, which stays at 90° while you drag: that is the third altitude passing through H.
Opening scene, frozen
The altitudes AD and BE meeting at the hollow point H.
Three Parallels
The construction draws, through each vertex, the line parallel to the opposite side: through A parallel to BC, through B parallel to CA, through C parallel to AB. The player marks each pair of parallels with the same chevrons, one, two or three. The three lines cross in pairs at three points, named A′ (opposite A), B′ and C′, and they form a triangle A′B′C′ that contains ABC.
The large triangle is similar to the small one and twice its size, with its sides parallel to the small one's; it is sometimes called the anticomplementary triangle. The next step shows the one property of it that the proof needs.
Three parallels, frozen
Through each vertex the parallel to the opposite side: the large triangle A′B′C′, dashed.
The Midpoints
Look at the quadrilateral ABA′C. Its sides AB and CA′ are parallel by construction, and so are AC and BA′, so it is a parallelogram and its opposite sides are equal: BA′=AC. The quadrilateral AC′BC is a parallelogram for the same reason, so C′B=AC. Therefore C′B=BA′, and B is the midpoint of the side C′A′ of the large triangle.
The same argument at the other two vertices shows that A is the midpoint of B′C′ and C the midpoint of A′B′. The player marks the six half-sides with one, two and three ticks. So the small triangle ABC is the triangle of midpoints of the large one, which is the midsegment theorem picture seen from the outside.
The midpoints, frozen
A, B, C are the midpoints of the sides of A′B′C′: one, two and three ticks.
Altitudes as Perpendicular Bisectors
Take the altitude BE. It is perpendicular to CA. The side C′A′ of the large triangle is parallel to CA, and a line perpendicular to one of two parallel lines is perpendicular to the other, so BE⊥C′A′. And BE passes through B, which is the midpoint of C′A′. A line through the midpoint of a segment, perpendicular to it, is its perpendicular bisector:
the line BE is the perpendicular bisector of C′A′
In the same way the altitude AD, perpendicular to BC and through the midpoint A of B′C′, is the perpendicular bisector of B′C′. The player marks the right angles at A and B between the altitude and the side of the large triangle.
Altitudes as perpendicular bisectors, frozen
Right angles at A and B between each altitude and the side of the large triangle.
The Circumcentre of the Big Triangle
The perpendicular bisectors of a triangle meet at one point, its circumcentre. Apply that to the large triangle A′B′C′. Two of its perpendicular bisectors are the lines AD and BE, and they meet at H. So H is the circumcentre of A′B′C′, and the third perpendicular bisector, that of the side A′B′, passes through H as well.
Nothing new was proved here; the work was done on the circumcentre page. What this page adds is the observation that the altitudes of one triangle are the perpendicular bisectors of another.
The circumcentre of the big triangle, frozen
AD and BE are two perpendicular bisectors of A′B′C′; they meet at H.
Why It Holds
The perpendicular bisector of A′B′ is the line through its midpoint, which is C, perpendicular to A′B′. Since A′B′∥AB, that is the line through C perpendicular to AB: the third altitude CF. It passes through H, so all three altitudes of ABC meet at H, the orthocentre. The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the given altitudes, a construction, parallelogram sides, perpendicularity carried across parallels, the circumcentre theorem, and the identification of the third line.
The proof gives a little more than concurrence: the orthocentre of a triangle is the circumcentre of its anticomplementary triangle. Turned round, the circumcentre of any triangle is the orthocentre of its medial triangle, the triangle of its midpoints.
The conclusion, frozen
The third altitude CF drawn: it passes through H.
Dragging the Figure
Drag a vertex and watch all three altitudes swing while still meeting at H; the readout keeps the angle between CH and AB at 90°. Pull A towards one side and H moves towards the opposite side; for an equilateral triangle H coincides with the centroid, the circumcentre and the incentre. The large triangle moves with the small one, always twice its size.
The player keeps the triangle acute, so that the feet of the altitudes lie on the sides and H stays inside; for a right triangle H is the vertex of the right angle, and for an obtuse triangle it lies outside, with two of the altitudes meeting the extensions of the sides. The theorem holds in all three cases. A drag that would carry a corner of the large triangle out of the frame is refused.
After dragging A, frozen
A pulled right: the altitudes lean and H has moved, still on all three of them.
Consequences
• Where it is. Inside an acute triangle, at the right-angle vertex of a right triangle, outside an obtuse triangle. • The Euler line. The orthocentre, the centroid and the circumcentre lie on one line, with the centroid between them at a third of the way from the circumcentre; this follows from the proof above, since the large triangle is a half-turn-and-double of the small one about the centroid. • An orthocentric system. Each of the four points A, B, C, H is the orthocentre of the triangle formed by the other three. • The orthic triangle. The feet D, E, F form a triangle whose angle bisectors are the altitudes of ABC, so H is its incentre; and A, F, H, E lie on a circle, since the angles at F and E are right, as in Thales' theorem and the cyclic quadrilateral page. • Reflections. The reflection of H in any side lies on the circumcircle of ABC.