# 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:

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