# Symmedian

A **symmedian** of a triangle is the reflection of the median on the bisector.

In the above diagram, $AM$ and $AD$ are median and symmedian of triangle $ABC$, respectively, with $\angle BAD = \angle CAM$.

## Drawing a Symmedian

The most popular way to draw a symmedian of a triangle is this:

Let $ABC$ be a triangle inscribed a circle with center $O$. Draw tangents of the circle $(O)$ at $B$ and $C$, they meet at point $P$. Then $AP$ is the symmedian of the triangle $ABC$.There is also another way of drawing a symmedian of a triangle:

Here is how we prove the way of drawing:

Let $AM$ be the median of triangle $ABC$. Let $D$ be the intersection of line $AP$ and the circle $(O)$. Let $E$ be the foot of perpendicular from $O$ to $AD$. Then $E$ is the midpoint of segment $AD$.

$\angle OEP = \angle OBP = 90^\circ$; so, $OEBP$ is a concylic quadrilateral. This gives us $\angle BEP = \angle BOP =\dfrac 12 \angle BOC = \angle BAC$. Also, $\angle BDA = \angle BCA$; so we have triangle $BED$ and triangle $BAC$ are similar. But $E$ and $M$ are midpoints of $AD$ and $BC$, respectively; hence, triangle $BDA$ and triangle $MCA$ are similar, or $\angle BAD = \angle MAC$.

Thus, $AD$ is the symmedian of triangle $ABC$.

From the upper two problems, we conclude to this result:Let $ABC$ be a triangle inscribed a circle with center $O$. Draw tangent of the circle $(O)$ at $A$, it cuts line $BC$ at point $K$. From $K$, draw tangent of the circle $(O)$ at point $D$. Then $AD$ is the symmedian of triangle $ABC$.

Here is how we prove the way of drawing:

Let $H$ be the intersection of $OK$ and $AD$. Because $KA$ and $KD$ are tangents from $K$ to the circle $(O)$, $OK$ is perpendicular to $AD$, and $H$ is the midpoint of segment $AD$.

$KB \times KC = KA^2$ (the power of point $K$). But $KA^2 = KH \times KO$; so, $KB \times KC = KH \times KO$, which makes $BHOC$ a concylic quadrilateral $\Rightarrow \angle KHB = \angle OCK$.

Now, $OH \times OK = OA^2 = OC^2 \Rightarrow \dfrac{OH}{OC}=\dfrac{OC}{OK}$. Hence, triangle $OHC$ is similar to triangle $OCK$; or, $\angle OHC = \angle OCK = \angle KHB \Rightarrow \angle BHD = \angle CHD = \dfrac12 \angle BHC$.

But, $\angle BHC = \angle BOC$ (property of concylic quadrilateral $BHOC$) $\Rightarrow \angle BHD = \dfrac12 \angle BOC = \angle BAC$. The last part is the same as the upper example.

Let $M$ be a point outside a circle with center $O$. Draw tangents $MA$ and $MB$ to $(O)$, and a line $MCD$ intersecting the circle $(O)$ at $C$ and $D$. Then line $AB$ and tangents of $(O)$ at $C$ and $D$ are concur.

## Calculations

Consider the following diagram: Triangle $ABC$ inscribed a circle with center $O$. $AD, AM$ are symmedian and median of the triangle, respectively (point $D$ lies on the circle). $AD$ intersects $BC$ at $N$. Let $BC = a, CA = b, AB = c$.

The goal is to calculate $NB, NC, AD$ and $AN$.

To calculate $NB$ and $NC$ we can use this result:

Let $ABC$ be a triangle. $M, N$ are points on $BC$ such that $\angle BAN = \angle CAM$. Then $\dfrac{NB}{NC}\times \dfrac{MB}{MC} = \dfrac{AB^2}{AC^2}$.If $AN, AM$ are symmedian and median of triangle $ABC$, respectively, then $\angle BAN = \angle CAM$. Use the upper result, we have: $\dfrac{NB}{NC} \times \dfrac{MB}{MC} = \dfrac{c^2}{b^2}$. But $\dfrac{MB}{MC}=1$, the equation changes to:

From $C$ draw a line parallel to $AB$, it intersects $AM, AN$ at $E, F$, respectively. Then $\angle CAM = \angle BAN = \angle CFA$, which gives us triangle $CEA$ and triangle $CAF$ similar $\Rightarrow CE \times CF = CA^2$.

By Intercept Theorem, $\dfrac{NB}{NC}=\dfrac{AB}{CF}$ and $\dfrac{MB}{MC}=\dfrac{AB}{CE}$. Multiply them together, we have: $\dfrac{NB}{NC} \times \dfrac{MB}{MC} = \dfrac{AB^2}{CE \times CF} = \dfrac{AB^2}{AC^2}$.

$\dfrac{NB}{NC}=\dfrac{c^2}{b^2}$

We write the equation in the form: $\dfrac{NB}{c^2}=\dfrac{NC}{b^2}=\dfrac{a}{b^2+c^2}$. Hence,

$NB=\dfrac{ac^2}{b^2+c^2}, NC=\dfrac{ab^2}{b^2+c^2}$

Now, to calculate $AD$, triangle $ABD$ and triangle $AMC$ are similar; so, $\dfrac{c}{AM}=\dfrac{AD}{b}$

$AD=\dfrac{bc}{AM}$

But, $AM=\dfrac{\sqrt{2(b^2+c^2)-a^2}}{2}$ (see centroid), the equation changes to:

$AD=\dfrac{2bc}{\sqrt{2(b^2+c^2)-a^2}}$

The last thing is to calculate $AN$. Let $AN = x$ then $DN = AD - x$.

$NA \times ND = NB \times NC$ (the power of point $N$) $\Rightarrow x(AD - x) = NB \times NC$.

Solve this quadratic equation, we have:

$AN = x = \dfrac{AD + \sqrt{AD^2 - 4NB \times NC}}{2}$