The Apollonius Theorem

Post 2: The Apollonius Theorem

Hello, I hope you all had a great day! In this post, we will be exploring the Apollonius Theorem, which can be thought of as a special case of the Stewart Theorem. The Apollonius Theorem will relate the length of the median of the triangle to the length of its sides. The Apollonius Theorem states that the sum of the squares of any two sides of any triangle equals twice the square on half the third side, together with twice the square on the median bisecting the third side. So, in other words, c^2+b^2=2(m^2+d^2) in the diagram below. Like the Stewart Theorem, we will prove this theorem using the law of cosines.

<Proof>

b^2=m^2+d^2+2dmcos𝜃  ….①

c^2=m^2+d^2+2dmcos𝜃’ (<= cos𝜃=cos𝜃’ since 𝜃 and 𝜃’ are supplements)

=m^2+d^2 - 2dmcos𝜃 ….②

① + ② yields

b^2+c^2=2(m^2+n^2), which is the Apollonius Theorem.

The Stewart Theorem

Post 1: The Stewart Theorem

Matthew Stewart (1717-1785) was a Scottish mathematician who is known for the publication of his best known work, Some General Theorems of Considerable use in the Higher Parts of Mathematics. This book is best known for “proposition II”, or what is now known as Stewart’s Theorem, which yields a relation between the lengths of the sides of the triangle and the length of a cevian of the triangle.

For those of us not familiar with “cevian,” a cevian is any line segment in a triangle that has one endpoint on a vertex of the triangle and the other on the opposite side. So, for example, medians, altitudes, and angle bisectors are all cevians. So what exactly is the Stewart Theorem?

Theorem: Suppose a, b and c are sides of a triangle. Let d be the length of a cevian to the side of length a. If the cevian divides a into two segments of length m and n, with m adjacent to c and n adjacent to b, then b^2m+c^2n=a(d^2+mn).  Notice, if you were to foil this out, you would get man+dad=bmb+cnc, which can be memorized easily with the mnemonic device, “A man and his dad put a bomb in a sink.”

Now, before we go about proving this theorem, are we all familiar with the law of cosines? If yes great, if not, don’t worry as I will be going over the law of cosines as the proof of this theorem can be seen as an application of the law of cosines.

So let us observe this triangle. The law of cosine states: a^2=b^2+c^2-2bccosA, b^2=a^2+c^2-2accosB, c^2=a^2+b^2-2abcosC

Now that we have reviewed the law of cosines, we will now observe the proof of the Stewart theorem.

In the given diagram, 𝛳 and 𝛳’ are supplementary. Therefore cos𝛳=cos𝛳’. From the law of cosines, we can write

b^2= n^2+d^2-2dncos𝛳’=n^2+d^2+2dncos𝛳 => ①

c^2=m^2+d^2-2mdcos𝛳 => ②

①* m +②* n=b^2m+c^2n=nm^2+n^2m+(n+m)d^2=(m+n)(mn+d^2) =a(mn+d^2), and this equation is the theorem.