Six questions about the centres of a triangle. Every one is a proof, and every one uses just three tools: the midpoint theorem, "a perpendicular bisector is the set of points equidistant from two ends", and "triangles with equal bases on one line and the same apex have equal areas". Read each question, try it in your notebook (use the hint if you are stuck), then press Next ▶ to watch the argument grow.
📖 Tools you will use
Midpoint theorem (page 8): a line through the midpoint of one side parallel to another side bisects the third side, and the segment is half the parallel side.
Perpendicular bisector (page 2): every point on it is equidistant from the two ends of the segment.
Areas (page 3): triangles with equal bases on the same line and the same apex have equal areas; a median halves the area of a triangle.
Centroid (page 10): the medians meet at G and AG : GD = 2 : 1. Orthocentre: the altitudes meet at H.
Question 1 · The midpoint of the hypotenuse
Q1Draw a right triangle and draw the perpendicular from the midpoint of the hypotenuse to the base. (i) Prove that this perpendicular is half the perpendicular side of the big triangle. (ii) Prove that the distances from the midpoint of the hypotenuse to the three vertices are equal. (iii) Prove that the circumcentre of a right triangle is the midpoint of the hypotenuse.
(ഒരു മട്ടത്രികോണം വരച്ച്, കർണ്ണത്തിന്റെ മധ്യബിന്ദുവിൽനിന്ന് പാദത്തിലേയ്ക്ക് ലംബം വരയ്ക്കുക …)
Call the triangle ABC with the right angle at B, M the midpoint of the hypotenuse AC and N the foot of the perpendicular from M to BC. MN and AB are both perpendicular to BC – what does that make them? Then the midpoint theorem does (i), and a perpendicular bisector does (ii).
🎬 Solution animation
Answer: (i) MN ∥ AB (both ⊥ BC); M is the midpoint of AC, so by the midpoint theorem N is the midpoint of BC and MN = ½AB. (ii) MN is the perpendicular bisector of BC ⇒ MB = MC; M is the midpoint of AC ⇒ MA = MC. Hence MA = MB = MC. (iii) M is equidistant from A, B and C, so M is the circumcentre: the circumcentre of a right triangle is the midpoint of the hypotenuse, and the circumradius is half the hypotenuse.
🖐️ Drag A and C – the circle through the three corners is centred at M
The right angle stays at B. Whatever the shape, the circle with centre M and radius MA passes through B as well: MA = MB = MC.
🌍 Why this matters
Thales knew this 2600 years ago in the form "an angle in a semicircle is a right angle". Carpenters use it backwards: to check that a corner is square, they measure from the corner to two marks on the arms and compare with the distance between the marks – the midpoint of the hypotenuse must be equidistant from all three points.
Question 2 · One point for all three centres
Q2Prove that in any equilateral triangle the circumcentre, the orthocentre and the centroid are the same point.
Take the median from one vertex. Because the two sides from that vertex are equal, the two halves of the triangle are congruent (SSS). What does that tell you about the angles at the midpoint?
🎬 Solution animation
Answer: Let AD be the median from A (D the midpoint of BC). ΔABD ≅ ΔACD (AB = AC, BD = DC, AD common – SSS), so ∠ADB = ∠ADC = 90°. Thus AD is also the altitude from A and the perpendicular bisector of BC. The same is true for the medians from B and C. So the three medians, the three altitudes and the three perpendicular bisectors are the same three lines, and their common point is at once the centroid, the orthocentre and the circumcentre.
Question 3 · Six equal areas
Q3The medians divide the triangle in the figure into six small triangles. Prove that all six have the same area.
Two small triangles that share the vertex G and have equal bases on the same side (for example AGF and BGF with AF = FB) have equal areas. Call the three pairs x, y, z and use the fact that a median halves the whole triangle.
🎬 Solution animation
Answer: Let the areas be AGF = BGF = x (equal bases AF = FB, same apex G), BGD = CGD = y, CGE = AGE = z. The median AD halves ΔABC: area ABD = area ACD ⇒ x + x + y = z + z + y ⇒ x = z. The median BE halves it too: area ABE = area CBE ⇒ x + x + z = y + y + z ⇒ x = y. So x = y = z: all six triangles have the same area, each ⅙ of ΔABC.
Question 4 · Three equal areas
Q4In the figure the centroid is joined to the three vertices, dividing the triangle into three small triangles. Prove that the three have the same area.
Each of the three triangles is made of two of the six triangles of Question 3.
🎬 Solution animation
Answer: Draw the medians. ΔGAB = AGF + BGF = 2x, ΔGBC = BGD + CGD = 2y, ΔGCA = CGE + AGE = 2z. By Question 3, x = y = z, so the three triangles have equal areas – each is ⅓ of ΔABC. (A shorter way: ΔGBC has base BC and height ⅓ of the height of ΔABC, because G is one third of the way up the median; so its area is ⅓ of the whole.)
Question 5 · The medial triangle shares the centroid
Q5The midpoints of the sides of the blue triangle are joined to form the small green triangle. The red line is a median of the big triangle. (i) Prove that this median bisects the top side of the small triangle. (ii) Prove that the centroid of the big triangle is also the centroid of the small triangle.
Name the big triangle ABC with D, E, F the midpoints of BC, AB, CA; the median is AD and it meets EF at X. Apply the midpoint theorem in the two halves ΔABD and ΔADC.
🎬 Solution animation
Answer: (i) EF joins midpoints, so EF ∥ BC. In ΔABD, E is the midpoint of AB and EX ∥ BD, so X is the midpoint of AD and EX = ½BD = ¼BC. In ΔADC, X is the midpoint of AD and XF ∥ DC, so XF = ½DC = ¼BC. Hence EX = XF: the median AD bisects EF. (ii) So the median of the small triangle DEF from D goes to X, which lies on AD – it is part of AD. In the same way the other two medians of DEF are parts of the medians BF and CE of ABC. The medians of DEF therefore meet where the medians of ABC meet: the two triangles have the same centroid.
Question 6 · The orthocentre of HBC
Q6In the figure H is the orthocentre of triangle ABC. Prove that the orthocentre of triangle HBC is A.
Write down what "H is the orthocentre" means: AH ⊥ BC, BH ⊥ CA, CH ⊥ AB. Now find the three altitudes of the new triangle HBC – each one is a line you already have.
🎬 Solution animation
Answer: Since H is the orthocentre of ABC: AH ⊥ BC, BH ⊥ CA and CH ⊥ AB. In ΔHBC: the altitude from H is the perpendicular to BC through H – that is the line HA. The altitude from B is the perpendicular to HC through B; since CH ⊥ AB, this is the line BA. The altitude from C is the perpendicular to HB through C; since BH ⊥ CA, this is the line CA. The three lines HA, BA, CA all pass through A, so A is the orthocentre of ΔHBC.
🖐️ Four points, each the orthocentre of the other three
Drag A, B or C. The six lines AB, BC, CA, HA, HB, HC come in three perpendicular pairs (AH ⊥ BC, BH ⊥ CA, CH ⊥ AB). Pick any three of the four points: the fourth is their orthocentre. Mathematicians call A, B, C, H an orthocentric system.
Check yourself
🧭 What this exercise taught
In a right triangle the midpoint of the hypotenuse is the circumcentre; the circumradius is half the hypotenuse.
In an equilateral triangle every median is also an altitude and a perpendicular bisector, so the three centres coincide.
The medians cut a triangle into six triangles of equal area; joining the centroid to the vertices gives three equal areas.
The medial triangle (midpoints joined) has the same centroid as the original triangle.
If H is the orthocentre of ABC, then A is the orthocentre of HBC (and B of HCA, C of HAB).