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Author |
hbghlyj
Posted 2023-3-30 00:25
FG201108
Theorem 3. A convex quadrilateral is subdivided into four nonoverlapping triangles by its diagonals. Consider the four tangency points of the incircles in these triangles on one of the diagonals. It is a tangential quadrilateral if and only if the distance between two tangency points on one side of the second diagonal is equal to the distance between the two tangency points on the other side of that diagonal.
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By the two tangent theorem we have
\begin{aligned}
& A T_1=A T_1''=A P-P T_1'', \\
& B T_1=B T_1'=B P-P T_1',
\end{aligned}so that
$$
A B=A T_1+B T_1=A P+B P-P T_1^{\prime\prime}-P T_1'.
$$
Since $P T_1^{\prime\prime}=P T_1'$
$$
A B=A P+B P-2 P T_1'
$$
In the same way
$$
C D=C P+D P-2 P T_3'
$$
Adding the last two equalities yields
$$
A B+C D=A C+B D-2 T_1'T_3'
$$
In the same way we get
$$
B C+D A=A C+B D-2 T_2'T_4'
$$
Thus
$$
A B+C D-B C-D A=-2\left(T_1'T_3'-T_2'T_4'\right)
$$
The quadrilateral has an incircle if and only if $A B+C D=B C+D A$. Hence it is a tangential quadrilateral if and only if
$$
T_1'T_3'=T_2'T_4' ⇔ T_1'T_2'+T_2'T_3'=T_2'T_3'+T_3'T_4' ⇔ T_1'T_2'=T_3'T_4'.
$$
Note that both $T_1'T_3'=T_2'T_4'$ and $T_1'T_2'=T_3'T_4'$ are characterizations of tangential quadrilaterals. It was the first of these two that was proved in [18].
![96_03-25[1].gif](data/attachment/forum/202303/29/185754kxj64jv4qv667vof.gif "96_03-25[1].gif")
Theorem 1. A convex quadrilateral is tangential if and only if the incircles in the two triangles formed by a diagonal are tangent to each other.
Proof. In a convex quadrilateral $A B C D$, let the incircles in triangles $A B C,C D A$, $B C D$ and $D A B$ be tangent to the diagonals $A C$ and $B D$ at the points $X,Y,Z$ and $W$ respectively (see Figure 2). First we prove that
$$
Z W=\frac{1}{2}|a-b+c-d|=X Y\text {. }
$$
Using the two tangent theorem, we have $B W=a-w$ and $B Z=b-z$, so
$$
Z W=B W-B Z=a-w-b+z .
$$
In the same way $D W=d-w$ and $D Z=c-z$, so
$$
Z W=D Z-D W=c-z-d+w .
$$
Adding these yields
$$
2 Z W=a-w-b+z+c-z-d+w=a-b+c-d .
$$
Hence
$$
Z W=\frac{1}{2}|a-b+c-d|
$$
where we put an absolute value since $Z$ and $W$ can "change places" in some quadrilaterals; that is, it is possible for $W$ to be closer to $B$ than $Z$ is. Then we would have $Z W=\frac{1}{2}(-a+b-c+d)$
The formula for $XY$ is derived in the same way.
Now two incircles on different sides of a diagonal are tangent to each other if and only if $XY=0$ or $ZW=0$. These are equivalent to $a+c=b+d$, which proves the theorem according to the Pitot theorem. |
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Ly-lie
Posted 2023-3-29 20:33
