# Medians and the Circle

Triangle △ABC is any arbitrary triangle. The medians of the triangle have been drawn (in green). The medians intersect the circumscribed circle at points P_{A}, P_{B}, and P_{C}. The medians intersect the opposite sides of the triangle at points A', B', and C'.

The altitudes of △CAP_{A} and △BAP_{A} have been drawn (in purple). Note that the areas of △BA'P_{A} and △CA'P_{A} are equal since P_{A}A' is a median of △BP_{A}C. Therefore, areas of △BAP_{A} and △CAP_{A} are equal. This means that the altitudes that have been drawn are equal in length.

Let AP_{A}, BP_{B}, and CP_{C} equal *N*_{A}, *N*_{B}, and *N*_{C}, respectively. We will find the length of these in terms of *a*, *b*, and *c*.

Knowing that the areas of △BAP_{A} and △CAP_{A} are equal, we can find an interesting relationship between *u* and *v*.

First, the areas of △BAP_{A} and △CAP_{A} can be given by $\frac{1}{2}cu\cdot \sin(\angle \mathrm{ABP_{A}})$ and $\frac{1}{2}bv\cdot \sin(\angle \mathrm{ACP_{A}})$, respectively. Angles ABP_{A} and ACP_{A} are opposite angles of a cyclic quadrilateral and are complementary. Therefore, the sine of these angles is equal. Therefore, we have the following relationships (and similarly with other sides):

(i) $\frac{b}{c} = \frac{u}{v}$, $\frac{a}{b}= \frac{y}{z}$, and $\frac{c}{a}= \frac{w}{x}$.

We can multiply all the relationships in (i) and obtain a curious relationship: $\frac{u}{v} \cdot \frac{w}{x} \cdot \frac{y}{z} = 1$.

Now, from the cyclic quadrilateral ABP_{A}C, we can state the following: $ub+vc = a\cdot N_{A}$. Since $u=\frac{b}{c}\cdot v$, this becomes $\frac{b^{2}}{c}\cdot v+vc = a\cdot N_{A}$ **(a)** $v(b^{2}+c^{2})= ac\cdot N_{A}$. Similarly, **(b)** $u(b^{2}+c^{2})= ab\cdot N_{A}$. The equalities in **(a)** and **(b)** are important later on.

Now, using the law of cosines on △ABP_{A} and △ACP_{A}, the following two relationships are true: $N_{A}^{2}=c^{2}+u^{2}-2uc\cdot \cos (\angle \mathrm{ABP_{A}})$ and $N_{A}^{2}=b^{2}+v^{2}-2vb\cdot \cos (\angle \mathrm{ACP_{A}})=b^{2}+v^{2}+2vb \cdot \cos(\angle \mathrm{ABP_{A}})$.

Substituting the value of cos(ABP_{A}) from one into the other, we have: $N_{A}^{2} = c^{2}+u^{2} - \frac{2uc}{2vb}\left [ N_{A}^{2} - (b^{2}+v^{2}) \right ]$. Since *uc* = *vb*, we have **(c)** $2N_{A}^{2}=b^2+c^2+u^2+v^2$.

Making the substitution from **(a)** and **(b)** into **(c)**, we have: $2N_{A}^{2}=b^2+c^2+\frac{a^2b^2N_{A}^{2}}{(b^2+c^2)^2}+\frac{a^2c^2N_{A}^{2}}{(b^2+c^2)^2}$ or **(d)** $2N_{A}^{2}=b^2+c^2+\frac{a^2N_{A}^{2}}{b^2+c^2}$. We can solve for the desired length *N*_{A} in **(d)**:

(ii) $N_{A}=\frac{b^2+c^2}{\sqrt{2(b^2+c^2)-a^2}}$.

We also note that the length of the median (this is the segment AA' and its length is denoted *m*_{A}) is given by:

(iii) $m_{A}=\frac{1}{2}\sqrt{2(b^2+c^2)-a^2}$

Therefore, the following is also true:

(iv) $m_{A}\cdot N_{A}=\frac{1}{2}(b^2+c^2)$

We also have the following relationships with *u* and *v* derived from **(a)**, **(b)**, and **(iv)**:

(v) $u=\frac{ab}{2m_{A}}$ and $v=\frac{ac}{2m_{A}}$. [When we divide *u* by *v* from these equalities, we obtain the original ratio of *b/c*.]

The areas of the triangles △BP_{A}C, △CP_{B}A, and △AP_{C}B are given by the formulas: $\frac{a^2K}{4m_{A}^2}$, $\frac{b^2K}{4m_{B}^2}$,
and $\frac{c^2K}{4m_{C}^2}$, respectively, where *K* is the area of △ABC.

The relationship in (iv) gives us an interesting inequality. This relationship can be stated as: $\frac{2m_{A}^{2}}{b^2+c^2}=\frac{m_{A}}{N_{A}}$ or **(e)** $\frac{2(b^2+c^2)-a^2}{4(b^2+c^2)}=\frac{1}{2}-\frac{a^2}{4(b^2+c^2)}=\frac{m_{A}}{N_{A}}$. Now, we know that **(f)** $0<\frac{m_{A}}{N_{A}}<1$. This can be deduced from the fact that there is not a triangle where *M*_{A} > *N*_{A} or *M*_{A} = *N*_{A}.

Therefore, we can rewrite **(e)** as: $0<\frac{1}{2}-\frac{a^2}{4(b^2+c^2)}<1$ or $0<2-\frac{a^2}{b^2+c^2}<4$ or

(vi) $\frac{a^2}{b^2+c^2}<2$.