Symmedian and bisectors meet at the diagonal. Let $A$, $B$, $C$, and $D$ be points on the same circle, and let the bisectors of the angles $\angle ABC$ and $\angle ADC$ intersect on the diagonal $AC$ at point $K$. Let $BD$ intersect $AC$ in $P$. Prove that $BD$ ($BP$ and $DP$) are symmedians for the triangles $\triangle ABC$ and $\triangle ADC$.
 A: Part of the answer below is copied (with edits) from my AoPS post #103771, which incidentally gives some context for this problem (it is
about harmonic quadrilaterals); see AoPS post #509134 by Virgil Nicula
for more of it.
We shall use the notation $\left\vert XY\right\vert $ for the unsigned length
of a segment $XY$, and we shall use the notation $\overline{XY}$ for the
signed length of a segment $XY$. (The latter length, of course, depends on an
orientation chosen on the line $XY$. However, ratios such as $\dfrac
{\overline{XY}}{\overline{YZ}}$ for three colinear points $X$, $Y$ and $Z$ are
independent of this orientation.)
Theorem 1. Let $ABC$ be a triangle. Let the tangent to the circumcircle of
triangle $ABC$ at the vertex $A$ meet the sideline $BC$ at the point $D$. Then,
$\dfrac{\overline{BD}}{\overline{DC}}=\dfrac{\left\vert AB\right\vert ^{2}
}{\left\vert CA\right\vert ^{2}}$.
Proof. We use non-directed angles and non-directed segments. The point $D$
either lies on the extension of the segment $BC$ beyond $C$, or on the
extension of the segment $CB$ beyond $B$. We shall assume that it lies on the
latter; the other case is analogous (some angles $\varphi$ are replaced by
$180^{\circ}-\varphi$) and is left to the reader.
As an angle between a tangent and a chord, the angle $\measuredangle DAB$ is
equal to the chordal angle $\measuredangle ACB$ of the chord $AB$. So we have
$\measuredangle DAB=\measuredangle ACB=\measuredangle DCA$. Furthermore,
$\measuredangle ADB=\measuredangle CDA$. Thus, the triangles $DAB$ and $DCA$
have two equal pairs of angles, and therefore are (oppositely) similar. Consequently,
(1) $\dfrac{\left\vert BD\right\vert }{\left\vert AD\right\vert }
=\dfrac{\left\vert AD\right\vert }{\left\vert CD\right\vert }=\dfrac
{\left\vert AB\right\vert }{\left\vert CA\right\vert }$.
Hence,
$\dfrac{\left\vert BD\right\vert }{\left\vert CD\right\vert }=\dfrac
{\left\vert BD\right\vert }{\left\vert AD\right\vert }\cdot\dfrac{\left\vert
AD\right\vert }{\left\vert CD\right\vert }=\dfrac{\left\vert AB\right\vert
}{\left\vert CA\right\vert }\cdot\dfrac{\left\vert AB\right\vert }{\left\vert
CA\right\vert }$ (here, we have applied (1) twice)
$=\dfrac{\left\vert AB\right\vert ^{2}}{\left\vert CA\right\vert ^{2}}$.
This is, so far, an equality between ratios of non-directed segments. However,
if we regard the segments on the left side as directed, then it is still
valid, because the point $D$ lies outside the segment $BC$ and therefore the
ratio $\dfrac{\overline{BD}}{\overline{CD}}$ is positive. Theorem 1 is thus proven.
Additionally, we state two well-known theorems:
Theorem 2. Let $ABC$ be a triangle. Let the angle bisector of the angle
$CAB$ meet the sideline $BC$ at the point $D$. Then,
$\dfrac{\overline{BD}}{\overline{CD}}=-\dfrac{\left\vert AB\right\vert
}{\left\vert CA\right\vert }$.
Theorem 3. Let $ABC$ be a triangle. Let the $A$-symmedian of triangle
$ABC$ meet the sideline $BC$ at the point $D$. Then,
$\dfrac{\overline{BD}}{\overline{CD}}=-\dfrac{\left\vert AB\right\vert ^{2}
}{\left\vert CA\right\vert ^{2}}$.
See http://www.cut-the-knot.org/triangle/symmedians.shtml for Theorem 3.
Theorem 4. Let $ABC$ be a triangle. Let the tangents to the circumcircle
of triangle $ABC$ at the points $B$ and $C$ intersect at $Q$. Then, $AQ$ is
the $A$-symmedian of triangle $ABC$.
See http://www.cut-the-knot.org/Curriculum/Geometry/Symmedian.shtml for the
proof of Theorem 4.
Theorem 5. Let $ABCD$ be a convex cyclic quadrilateral with the
circumcircle $k$. Then, the following three statements are equivalent:
Statement 1: The tangents to $k$ at the vertices $A$ and $C$ intersect on
the diagonal $BD$.
Statement 2: The tangents to $k$ at the vertices $B$ and $D$ intersect on
the diagonal $AC$.
Statement 3: We have $\left\vert AB\right\vert \cdot\left\vert CD\right\vert
=\left\vert BC\right\vert \cdot\left\vert DA\right\vert $.
Statement 4: The angle bisectors of angles $BCD$ and $BAD$ intersect on the
diagonal $BD$.
Statement 5: The angle bisectors of angles $ABC$ and $ADC$ intersect on the
diagonal $AC$.
Statement 6: The $A$-symmedian of triangle $DAB$ and the $D$-symmedian of
triangle $DCB$ intersect on the diagonal $BD$.
Statement 7: The $B$-symmedian of triangle $ABC$ and the $D$-symmedian of
triangle $ADC$ intersect on the diagonal $AC$.
Statement 8: The line $AC$ is the $A$-symmedian of triangle $DAB$ and is the
$D$-symmedian of triangle $DCB$.
Statement 9: The line $BD$ is the $B$-symmedian of triangle $ABC$ and is the
$D$-symmedian of triangle $ADC$.
(Of course, $BD$ and $AC$ are understood as lines, not as segments. When we
say that two lines $g$ and $g^{\prime}$ intersect on a line $h$, we allow $g$
and $g^{\prime}$ to coincide.)
Quadrilaterals $ABCD$ satisfying the nine equivalent statements of Theorem 5
are said to be harmonic.
Proof of Theorem 5. Let the tangent to $k$ at the vertex $A$ meet the
diagonal $BD$ at a point $A^{\prime}$, and let the tangent to $k$ at the
vertex $C$ meet the diagonal $BD$ at a point $C^{\prime}$. Then, the tangents
to $k$ at the vertices $A$ and $C$ intersect on the diagonal $BD$ if and only
if $A^{\prime}=C^{\prime}$. But when do we have $A^{\prime}=C^{\prime}$ ?
We shall use directed segments.
Theorem 1 (applied to $A$, $B$, $D$ and $A^{\prime}$ instead of $A$, $B$, $C$
and $D$) shows that $\dfrac{\overline{BA^{\prime}}}{\overline{DA^{\prime}}
}=\dfrac{\left\vert AB\right\vert ^{2}}{\left\vert DA\right\vert ^{2}}$.
Theorem 1 (applied to $C$, $B$, $D$ and $A^{\prime}$ instead of $A$, $B$, $C$
and $D$) shows that $\dfrac{\overline{BC^{\prime}}}{\overline{DC^{\prime}}
}=\dfrac{\left\vert CB\right\vert ^{2}}{\left\vert DC\right\vert ^{2}}$.
Now, we have the following logical equivalence:
$\left(  \text{Statement 1 holds}\right)  $
$\Longleftrightarrow\left(  \text{the tangents to }k\text{ at the vertices
}A\text{ and }C\text{ intersect on the diagonal }BD\right)  $
$\Longleftrightarrow\left(  A^{\prime}=C^{\prime}\right)  $
(because we have seen that the tangents to $k$ at the vertices $A$ and $C$
intersect on the diagonal $BD$ if and only if $A^{\prime}=C^{\prime}$)
$\Longleftrightarrow\left(  \text{the points }A^{\prime}\text{ and }C^{\prime
}\text{ divide the segment }BD\text{ in the same (signed) ratio}\right)  $
$\Longleftrightarrow\left(  \dfrac{\overline{BA^{\prime}}}{\overline
{DA^{\prime}}}=\dfrac{\overline{BC^{\prime}}}{\overline{DC^{\prime}}}\right)
$
$\Longleftrightarrow\left(  \dfrac{\left\vert AB\right\vert ^{2}}{\left\vert
DA\right\vert ^{2}}=\dfrac{\left\vert CB\right\vert ^{2}}{\left\vert
DC\right\vert ^{2}}\right)  $
(since $\dfrac{\overline{BA^{\prime}}}{\overline{DA^{\prime}}}=\dfrac
{\left\vert AB\right\vert ^{2}}{\left\vert DA\right\vert ^{2}}$ and
$\dfrac{\overline{BC^{\prime}}}{\overline{DC^{\prime}}}=\dfrac{\left\vert
CB\right\vert ^{2}}{\left\vert DC\right\vert ^{2}}$)
$\Longleftrightarrow\left(  \left\vert AB\right\vert ^{2}\cdot\left\vert
DC\right\vert ^{2}=\left\vert CB\right\vert ^{2}\cdot\left\vert DA\right\vert
^{2}\right)  $
$\Longleftrightarrow\left(  \left\vert AB\right\vert ^{2}\cdot\left\vert
CD\right\vert ^{2}=\left\vert BC\right\vert ^{2}\cdot\left\vert DA\right\vert
^{2}\right)  $
$\Longleftrightarrow\left(  \left\vert AB\right\vert \cdot\left\vert
CD\right\vert =\left\vert BC\right\vert \cdot\left\vert DA\right\vert \right)
$
$\Longleftrightarrow\left(  \text{Statement 3 holds}\right)  $.
Thus, we have proven the equivalence between Statement 1 and Statement 3.
Similarly, we can prove the equivalence between Statement 2 and Statement 3.
Hence, Statements 1, 2 and 3 are equivalent. A similar argument (but using
Theorem 2 instead of Theorem 1) proves that Statements 4, 5 and 3 are
equivalent. A similar argument (but using Theorem 3 instead of Theorem 1)
proves that Statements 6, 7 and 3 are equivalent. Thus, Statements 1, 2, 3, 4,
5, 6 and 7 are equivalent.
Clearly, Statement 9 implies Statement 7. Now we shall prove that Statement 7
implies Statement 9:
Assume that Statement 7 holds. Thus, Statement 1 holds (since we know that
Statements 1, 2, 3, 4, 5, 6 and 7 are equivalent). In other words, the
tangents to $k$ at the vertices $A$ and $C$ intersect on the diagonal $BD$.
Let $S$ be the point on $BD$ in which they intersect. Thus, the tangents to
the circumcircle of triangle $BCA$ at the points $C$ and $A$ intersect at $S$.
Theorem 4 (applied to $B$, $C$, $A$ and $S$ instead of $A$, $B$, $C$ and $Q$)
thus shows that $BS$ is the $B$-symmedian of triangle $BCA$. In other words,
$BS$ is the $B$-symmedian of triangle $ABC$. In other words, $BD$ is the
$B$-symmedian of triangle $ABC$ (since the line $BS$ is the line $BD$).
Similarly, the line $BD$ is the $D$-symmedian of triangle $ADC$. Thus,
Statement 9 holds.
Now, let us forget that we have assumed that Statement 7 holds. We thus have
shown that Statement 7 implies Statement 9. Since Statement 9 also implies
Statement 7, this shows that Statements 7 and 9 are equivalent. Similarly,
Statements 6 and 8 are equivalent. Combining this with the already-known
equivalence of Statements 1, 2, 3, 4, 5, 6 and 7, we conclude that all nine
Statements 1, 2, 3, 4, 5, 6, 7, 8 and 9 are equivalent. Theorem 5 is proven.
Your claim is part of Theorem 5: You are saying that Statement 5 implies
Statement 9.
