Apollonius’ Theorem gives a link between the three sides of any triangle and the median drawn to one of them. Take a triangle ABC and let point D be the midpoint of the side BC. The segment AD is called the median to BC because it cuts that side into two equal pieces. The theorem says that if you square the lengths of the other two sides, AB and AC, and then add those squares together, the result equals twice the square of the median AD plus half the square of the side BC. It holds for all triangles.

What is Apollonius Theorem?

Apollonius’ Theorem tells us how the lengths of these four segments—AB, AC, BC and the median AD—fit together. Square the lengths of the two sides that meet at A (that is, AB² and AC²) and add those two numbers. Now find the length of BC, cut it in half, square that half‑length, and multiply by two. Do the same—square and double—for the median AD. The theorem says the total you started with (AB² + AC²) will match the combined total you just worked out: 2 × (half of BC)² plus 2 × (AD)².

Or, we can say that for a triangle ABC where AM is the median bisecting side BC, according to Apollonius theorem,

\(AB^2+AC^2=2\left\{AM^2+\left(\frac{BC}{2}\right)^2\right\}\)

Apollonius Theorem Proof

We can prove Apollonius Theorem using various methods given below:

1. Apollonius Theorem Proof by Pythagoras Theorem
2. Apollonius Theorem Proof by Cosine Rule
3. Apollonius Theorem Proof by Vectors

Apollonius Theorem Proof by Pythagoras Theorem

Let ABC be the given triangle, we need to prove that triangle ABC with M as the midpoint of BC satisfies Apollonius theorem using Pythagoras theorem:

Let AH be the altitude of triangle ABC, that is H is the foot of the perpendicular from A to BC.

Also, as AM is the median, so M is the midpoint of BC.

BM = CM = BC/2

Or, BM + CM = BC

By Using Pythagoras Theorem, in triangle ABH, AMH and triangle ACH we get,

\( AB^2\ =\ BH^2+AH^2\)

\(AC^2\ =\ AH^2+CH^2\)

\(AM^2\ =\ AH^2+MH^2\)

Using the above three results we can conclude that:

\(AB^2+AC^2\ =\ 2AH^2+BH^2+CH^2\)

\(=\ 2AH^2+2MH^2+BH^2-MH^2+CH^2-MH^2\)

\(=\ 2AM^2+\left(BH+MH\right)\left(BH-MH\right)+\left(CH+MH\right)\left(CH-MH\right)\)

\(=\ 2AM^2+\left(BH+MH\right).BM+CM.\left(CH-MH\right)\)

\(=\ 2AM^2+\frac{BC^2}{2}\)

\(=\ 2\left\{AM^2+\left(\frac{BC}{2}\right)^2\right\}\)

Hence, we have proved the Apollonius theorem by Pythagoras theorem.

Apollonius Theorem Proof by Cosine Rule

Let ABC be the given triangle, we need to prove that triangle ABC with M as the midpoint of BC satisfies Apollonius theorem using the cosine rule:

Using the cosine rule, we get,

\(AB^2=BM^2+AM^2-2AM.BM\cos\angle AMB\)

\(AC^2=\ CM^2+AM^2-2AM.CM\cos\angle AMC\)

\(=\ BM^2+AM^2-2AM.BM\cos\angle AMB\), as \(\angle AMB+\angle AMC\ =\pi\)

Adding the above two equations:

\(AB^2+AC^2=2AM^2+2BM^2\)

\(=2\left\{AM^2+\left(\frac{BC}{2}\right)^2\right\}\)

Hence, we have proved the Apollonius Theorem by Cosine Rule.

Apollonius Theorem Proof by Vectors

Let ABC be the given triangle, we need to prove that triangle ABC with M as the midpoint of BC satisfies Apollonius theorem using elementary vector operations:

Consider A at the origin of the cartesian plane.

Let \(\vec{AB}=\vec{b}\) and \(\vec{AC}=\vec{c}\)

Then, \(\vec{BC}=\vec{c}-\vec{b}\), and \(\vec{BC}=\vec{c}-\vec{b}\)

So, \(AB^2+AC^2=\left|\vec{b}\right|^2+\left|\vec{c}\right|^2\)

\(=\frac{1}{2}\left(2\left|\vec{b}\right|^2+2\left|\vec{c}\right|^2\right)\)

\(=\frac{1}{2}\left(\left|\vec{b}\right|^2+\left|\vec{c}\right|^2+2\vec{b}.\vec{c}+\left|\vec{b}\right|^2+\left|\vec{c}\right|^2-2\vec{b}.\vec{c}\right)\)

\(=\frac{1}{2}\left\{\left(\vec{b}+\vec{c}\right)^2+\left(\vec{c}-\vec{b}\right)^2\right\}\)

\(=\frac{1}{2}\left(4AM^2+BC^2\right)\)

\(=2\left\{AM^2+\left(\frac{BC}{2}\right)^2\right\}\)

Hence, we have proved the Apollonius Theorem by Vectors.

Relation of Apollonius Theorem with other Theorems

We can see that Apollonius Theorem is a special case of Stewart’s Theorem and can be considered a generalization of Pythagoras Theorem.

Substitute BP = CP in Stewart’s Theorem, and we get:

\(=CP.AB^2+BP+AC^2=\left(BP+CP\right)\left(AP^2+BP.CP\right)\), and you get

\(=AB^2+AC^2=2\left(AP^2+BP^2\right)\)

Now, substituting \(\angle BAC=\frac{\pi}{2}\) and \(AM=\frac{BC}{2}\) into Apollonius Theorem:

\(AB^2+AC^2=2\left\{\left(\frac{BC}{2}\right)^2+\left(\frac{BC}{2}\right)^2\right\}\)

                       =\(BC^2\).

Properties of Apollonius Theorem

Applies to All Triangles:
Apollonius Theorem works for any triangle — whether it is scalene, isosceles, or equilateral.

Involves a Median:
The theorem relates the two sides of a triangle to the median drawn from the opposite vertex to the midpoint of the third side.

Mathematical Statement:
If A, B, and C are vertices of a triangle and D is the midpoint of side BC, then:
AB² + AC² = 2AD² + (1/2)BC²

Shows Geometric Balance:
The theorem shows how the sum of the squares of two sides is balanced by the length of the median and the third side.

Useful in Coordinate Geometry:
It helps in finding unknown side lengths or medians in coordinate-based problems.

Can Be Proved in Multiple Ways:
The theorem can be proven using the Pythagoras Theorem, Cosine Rule, or Vectors.

Helps in Median Length Calculation:
If the side lengths are known, the theorem can be used to find the length of the median to a side.

Basis for Further Theorems:
Apollonius Theorem is foundational and is often used in proofs and applications in higher-level geometry and physics.

Example 1: The sides of a triangle are 7cm, 6cm, and 10 cm respectively. What will be the length of the median to the side 10cm.

Solution: Let us suppose a = 10 cm, b = 7 cm and c = 6 cm.

As we have to find the length of the median to side 10 cm, let the length of the median be ‘d’.

Also, as the median on the side bisects the side, so, m = a/2 = 5 cm.

According to the apollonius theorem:

\(c^2+b^2= 2(m^2+d^2)\)

On substituting the values we get:

\(6^2+7^2= 2(5^2+d^2)\)

On further solving, we get:

\(36+49= 2(25+d^2)\)

\(85 – 50 = 2d^2\)

\(2d^2 = 35\)

\(d = \sqrt{ \frac {35}{2}}\)

d = 4.183 cm

Therefore, the length of the median to the side 10 cm of the given triangle is 4.183 cm.

Example 2: The sides of the parallelogram are 6 cm and 4 cm. The length of one of the diagonal of the parallelogram is 8 cm, find the length of the second diagonal.

Solution: We can compare the situation mentioned above as that of a triangle. Two sides and one diagonal of the parallelogram forms the three sides of a triangle.

The second diagonal of the parallelogram forms the median of the first diagonal. Both the diagonals of the parallelogram bisect each other.

Basically, we need to find the length of the median. Twice the length of the median will give us the length of the diagonal for the parallelogram.

Let us take the sides as:

a = 8 cm, b = 6 cm, and c = 4 cm, here m = a/2 = 4 cm.

According to the apollonius theorem:

\(c^2+b^2= 2(m^2+d^2)\)

\(4^2+6^2= 2(4^2+d^2)\)

\(16 + 36= 2(16 +d^2)\)

\( 52= 32 + 2d^2\)

\( 52 – 32 = 2d^2\)

\( 2d^2 = 20\)

\(d = \sqrt{10}\)

d = 3.16 cm

As already discussed, length of the diagonal is twice the length of the median:

So, Diagonal = 2d = 2(3.16) cm

Diagonal = 6.32 cm

Therefore, the length of the second diagonal for the given parallelogram is 6.32 cm.

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