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 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 \(ABC\).
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 \(ABC\).
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.
(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 \(D\). \(_\square\)
A triangle \(ABC\) is inscribed inside a circumcircle with center \(O\). Draw the median \(AM\) of the circle. Now, draw the tangents of the circle \(O\) at \(B\) and \(C\), and they cut at point \(P\). \(AP\) cuts the circle \(O\) at \(D\). If \(\angle BDM = 27^\circ\), find \(\angle AOC\) in degrees.
Clarification: For anyone who doesn't know what symmedian is, it's just the reflection of the median on bisector. For example, for triangle \(ABC\) with median \(AM\) and bisector \(AI,\) if you draw the reflection \(AD\) of \(AM\) on \(AI,\) then \(AD\) is the symmedian.
Calculations
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 \(\frac{NB}{NC}\times \frac{MB}{MC} = \frac{(AB)^2}{(AC)^2}\).
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
\[AN = x = \dfrac{AD + \sqrt{AD^2 - 4NB \times NC}}{2}.\]