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Prove that if a middle lane of a quadrangle is equal to half the sum of its sides, then the quadrangle is a trapezoid, i.e. given a quadrangle $\mathrm{ABCD}$ and the middle of $\mathrm{AB}$ is $\mathrm{H}$, the middle of $\mathrm{CD}$ is $\mathrm{K}$. Then if HK is $1 / 2$ of $B C+A D$, then $A B C D$ is a trapezoid, i.e. $B C$ is parallel to $A D$

Definition: A trapezoid (Figure 2 ) is a quadrilateral with two sides parallel. The middle lane is the line segment joining the middle points of two nonparallel sides.

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Proof: Assume $A B C D$ is not a trapezoid, i.e. $A D, H K$ and $B C$ are not parallel. Then, we can draw $A D^{\prime} / / \mathrm{HK}$ and $B C^{\prime} / /$ HK. Extend $A D^{\prime}$ such that $D^{\prime} S=B C^{\prime}$. And connect DS.

From $A D^{\prime} / / H K, B C^{\prime} / / H K$ and $H$ is the midpoint of $A B$, we know $A B C^{\prime} D^{\prime}$ is a trapezoid, so

$$\mathrm{HK}=\frac{1}{2}\left(\mathrm{AD}^{\prime}+\mathrm{BC}^{\prime}\right)$$

(5.1)

and $D^{\prime} K=C^{\prime} K$.

It is given that

$$\mathrm{HK}=\frac{1}{2}(\mathrm{AD}+\mathrm{BC})$$

$(5.2)$

and $D K=C K$.

And $D D^{\prime}=D K-D^{\prime} K, C C^{\prime}=C K-C^{\prime} K \cdot S O D D^{\prime}=C C^{\prime}$ '

From AD' // HK // BC', \&DD'A =<KC 'B $.$ So $<D D^{\prime} S=<C C^{\prime} B$, because they are supplementary angles of $\angle D D^{\prime} A$ and $<K C^{\prime} B$ respectively.

Now we know $D D^{\prime}=C C^{\prime}, D^{\prime} S=C^{\prime} B$ and $<D D^{\prime} S=<C C^{\prime} B$, so Triangle $D D^{\prime} S$ is congruent to Triangle CC'B. Hence $\mathrm{BC}=\mathrm{DS}$.

From (5.1) and (5.2), we get

$A D^{\prime}+B C^{\prime}=A D+B C$

By the congruence of $B C^{\prime}=S D^{\prime}$ and $B C=D S$, we have

$A D^{\prime}+S D^{\prime}=A D+D S$

i.e.

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$A S=A D+D S$

This can happen only if $A, D, S$ are on a line, that means $A D / / H K / / B C .$ So $A B C D$ is a trapezoid.

by Platinum (131,378 points)

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