4
$\begingroup$

Theorem

The midpoints of the bases of a trapezoid and the intersection of the diagonals of the same trapezoid are collinear.

Demonstration

$\mathit{ABCD}$ is a trapezoid, and $\mathit{AB}$ and $\mathit{CD}$ are its bases. $O$ is the intersection of the diagonals, and $P$ is the intersection of the lines through the legs of the trapezoid. The line through $O$ and $P$ intersects the bases $\overline{\mathit{AB}}$ and $\overline{\mathit{CD}}$ at $M$ and $N$, respecitvely. $\triangle\mathit{APM} \sim \triangle\mathit{DPN}$, and \begin{equation*} \frac{\mathit{AM}}{\mathit{DN}}=\frac{\mathit{PM}}{\mathit{PN}} . \end{equation*} $\triangle\mathit{BPM} \sim \triangle\mathit{CPN}$, and \begin{equation*} \frac{\mathit{PM}}{\mathit{PN}} = \frac{\mathit{BM}}{\mathit{CN}} . \end{equation*} So, \begin{equation*} \frac{\mathit{AM}}{\mathit{DN}} = \frac{\mathit{BM}}{\mathit{CN}} . \end{equation*} Likewise, since $\triangle\mathit{AOM} \sim \triangle\mathit{CON}$, and $\triangle\mathit{BMO} \sim \triangle\mathit{DNO}$, \begin{equation*} \frac{\mathit{AM}}{\mathit{CN}} = \frac{\mathit{BM}}{\mathit{DN}} . \end{equation*} I would like to show that the four points $M$, $N$, $O$, and $P$ are collinear. (I will respond to this post with a TikZ diagram.)

$\endgroup$
  • $\begingroup$ What's your question? $\endgroup$ – Ng Chung Tak Mar 5 '17 at 19:58
  • $\begingroup$ @Ng Chung Tak From the argument that I have presented, show that $M$, $N$, $O$, and $P$ are collinear. $\endgroup$ – user366533 Mar 5 '17 at 21:45
1
$\begingroup$

Here is a TikZ diagram to accompany the argument provided.

\begin{tikzpicture}


%Trapezoid ABCD is drawn. The lines through its legs intersect at P.  The intersection
%of the diagonals is labeled O. The intersection of the line through O and P
%intersects the bases at M and N.
\coordinate (A) at (0,0);
\node[anchor=north, inner sep=0, font=\footnotesize] at (0,-0.15){$A$};
\coordinate (B) at (6,0);
\node[anchor=north, inner sep=0, font=\footnotesize] at ($(B) +(0,-0.15)$){$B$};
\coordinate (C) at ($(B) +(-2.5,4)$);
%\node[anchor=north, inner sep=0, font=\footnotesize] at ($(C) +(0,-0.15)$){$C$};
\coordinate (D) at ($(A) +(1,4)$);
%\node[anchor=north, inner sep=0, font=\footnotesize] at ($(D) +(0,-0.15)$){$D$};
%
\draw (A) -- (B) -- (C) -- (D) -- cycle;
%
%Lines through the legs of the trapezoid are drawn.
%
\path[name path=a_path_to_locate_P] let \p1=($(A)-(D)$), \n1={atan(\y1/\x1)} in (D) -- ($(D) +(\n1:3.25)$);
\path[name path=another_path_to_locate_P] let \p1=($(B)-(C)$), \n1={atan(\y1/\x1)} in (C) -- ($(C) +({\n1+180}:3.75)$);
%
\coordinate[name intersections={of=a_path_to_locate_P and another_path_to_locate_P, by=P}];
%
\draw[dashed] (P) -- (C);
\draw[dashed] (P) -- (D);
%
%
%The labels for the vertices of the trapezoid are typeset.
\path[name path=a_path_to_typeset_C] ($(B)!0.15cm!-90:(P)$) -- ($(P)!0.15cm!90:(B)$);
\path[name path=another_path_to_typeset_C] (C) -- ($(C)!-0.2cm!(D)$);
\coordinate[name intersections={of=a_path_to_typeset_C and another_path_to_typeset_C, by=label_C}];
\node[anchor=west, inner sep=0, font=\footnotesize] at (label_C){$C$};
\path[name path=a_path_to_typeset_D] ($(A)!0.15cm!90:(P)$) -- ($(P)!0.15cm!-90:(A)$);
\path[name path=another_path_to_typeset_D] (D) -- ($(D)!-0.2cm!(C)$);
\coordinate[name intersections={of=a_path_to_typeset_D and another_path_to_typeset_D, by=label_D}];
\node[anchor=east, inner sep=0, font=\footnotesize] at (label_D){$D$};
%
%
%The label for P is typeset.
\coordinate (M) at ($(A)!0.5!(B)$);
\coordinate (N) at ($(C)!0.5!(D)$);
\draw let \p1=($(M)-(N)$), \n1={atan(\y1/\x1)} in node[anchor=\n1, inner sep=0, font=\footnotesize] at ($(P) +({\n1-180}:0.15)$){$P$};


%The line segment from P through M and N is drawn.
\draw[dashed] (P) -- (M);


%The diagonals of the trapezoid are drawn.
\draw[name path=a_path_to_locate_O, dashed] (A) -- (C);
\draw[name path=another_path_to_locate_O, dashed] (B) -- (D);
%
\coordinate[name intersections={of=a_path_to_locate_O and another_path_to_locate_O, by=O}];


%Labels for the midpoints of the bases of the trapezoid are typeset.
\draw let \p1=($(M)-(N)$), \n1={atan(\y1/\x1)} in node[anchor=south west, inner sep=0, font=\footnotesize] at ($(M) +({0.5*(\n1+180)}:0.15)$){\textit{M}};
\draw let \p1=($(M)-(N)$), \n1={atan(\y1/\x1)} in node[anchor=south west, inner sep=0, font=\footnotesize] at ($(N) +({0.5*(\n1+180)}:0.15)$){\textit{N}};


%A "pin" is drawn to M and N.
\draw[draw=gray, shorten <=1mm, shorten >=1mm] (O) -- ($(O) +(0.5,0)$);
\node[anchor=west, inner sep=0, font=\footnotesize] at ($(O) +(0.5,0)$){\textit{O}};

\end{tikzpicture}
| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Construction: Through O, draw HOK // AB cutting AD and BC at H and K respectively.

enter image description here

Besides the quoted theorem, there is another:-

With the given conditions, we should have HO = KO.


Let us assume that MO extended cuts AP at X and BP extended at P’.

Considering similar ⊿s P’OK and P’MB, we have $\dfrac {P’B}{P’K} =\dfrac {P’M}{P’Q} =\dfrac {MB}{OK}$

Similarly, we have $\dfrac {XA}{XH} =\dfrac {XM}{XO} =\dfrac {AM}{HO} = … =\dfrac {MB}{OK}$

$∴ \dfrac {p’M – P’O}{P’O} =\dfrac {XM - XO}{XO} \rightarrow \dfrac {OM}{P’M} =\dfrac {OM}{XM} = \rightarrow P’M = XM$. This means P’, X, P are actually the same point.

Result follows by combining such finding with the mentioned theorem.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy