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 $\angle BAD = \angle CAM$.

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

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

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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 $\angle BEP = \angle BOP =\frac 12 \angle BOC = \angle BAC$. 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 $ABC$. $_\square$

There is also another way of drawing a symmedian of a triangle:

Let $ABC$ be a triangle inscribed in a circle with center $O$. Draw tangent of the circle at $A$, and it cuts line $BC$ at point $K$. From $K$, draw tangent of the circle at point $D$. Then $AD$ is the symmedian of triangle $ABC$.

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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, $OK$ is perpendicular to $AD$, and $H$ is the midpoint of segment $AD$.

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

Now, $OH \times OK = (OA)^2 = (OC)^2 \implies \frac{OH}{OC}=\frac{OC}{OK}$. Hence, triangle $OHC$ is similar to triangle $OCK,$ or $\angle OHC = \angle OCK = \angle KHB \implies \angle BHD = \angle CHD = \frac12 \angle BHC$.

But, $\angle BHC = \angle BOC$ $($property of concylic quadrilateral $BHOC)$ $\implies \angle BHD = \frac12 \angle BOC = \angle BAC$. The last part is the same as the above example. $_\square$

From the above two problems, we conclude as follows:

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

Let $ABC$ be a triangle inscribed in a circle with center $O$ and a median $AM$. Let $(I_1)$ be a circle that passes through $B$ and $M$ and is tangent to line $AB$. Let $(I_2)$ be a circle that passes through $C$ and $M$ and is tangent to line $AC$. Prove that $(O), (I_1)$, and $(I_2)$ concur.

Source: Team Selection Test of High School for the Talented, HCMC, Vietnam.

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(You might be surprised that the intersection of the three circles is somewhat related to symmedian.)

Draw the symmedian $AD$ of triangle $ABC$, with point $D$ lying on the circle $(O)$. Extend $AM$ to intersect $(O)$ at point $E$. Because $AD$ and $AM$ are symmedian and median of triangle $ABC$, respectively, $\angle BAD = \angle CAE \implies \stackrel \frown{BD} = \stackrel \frown{CE}$. Hence, $BDEC$ is an isosceles trapezium. This gives us $BD = CE$ and $\angle CBD = \angle BCE$.

Now, have a look at triangle $BDM$ and $CEM$. It is quite obvious that they are congruent. So, $\angle BDM = \angle AEC = \angle ABC$, or $AB$ is tangent to the circumcircle of $BDM$.

But, $(I_1)$ is the circle that passes through $B, M$ and is tangent to $AB$, which means, $(I_1)$ is the circumcircle of $BDM;$ or $(I_1)$ passes through point $D$.

Thus, $(I_2)$ also passes through point $D$. We have already proved that the three circles concur at $D$. $_\square$

Consider the following diagram: triangle $ABC$ inscribed in a circle with center $O$. $AD$ and $AM$ 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 $\frac{NB}{NC}\times \frac{MB}{MC} = \frac{(AB)^2}{(AC)^2}$.

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

By intercept theorem, $\frac{NB}{NC}=\frac{AB}{CF}$ and $\frac{MB}{MC}=\frac{AB}{CE}$. Multiplying them together, we have $$\dfrac{NB}{NC} \times \dfrac{MB}{MC} = \dfrac{AB^2}{CE \times CF} = \dfrac{AB^2}{AC^2}.\ _\square$$

If $AN, AM$ are the symmedian and median of triangle $ABC$, respectively, then $\angle BAN = \angle CAM$. Using the above result, we have $\frac{NB}{NC} \times \frac{MB}{MC} = \frac{c^2}{b^2}$. But $\frac{MB}{MC}=1$, so the equation changes to

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

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

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

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

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

But, $AM=\frac{\sqrt{2(b^2+c^2)-a^2}}{2}$ (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 $DN = AD - x$. So,

$$NA \times ND = NB \times NC\ (\text{the power of point } N) \implies x(AD - x) = NB \times NC.$$

Solving this quadratic equation, we have

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

• Symmedian Point of a Triangle

• Isogonal Line