Tõesta, et trapetsi diagonaalide keskpunkte
ühendav surglõik on paralleelne selle alustega

Olgu M ja N trapetsi ABCD diagonaalide keskpunktid.

Näitame, et  MN ∥ AD. Selleks piisab, kui näidata, et   
         on kolineaarne        . Kuna M ja N on lõikude AC ja BD keskpunktid, siis

\overrightarrow{MN} 

\overrightarrow{AC}

\overrightarrow{AM}=\frac{1}{2}\overrightarrow{AC}=\frac{1}{2}\left(\overrightarrow{AB} +\overrightarrow{BC} \right),  

\overrightarrow{AN}=\frac{1}{2}\left( \overrightarrow{AB} +\overrightarrow{AD}  \right), 

\overrightarrow{MN}=\overrightarrow{AN}-\overrightarrow{AM}=\frac{1}{2}\left(  \overrightarrow{AB}+\overrightarrow{AD} \right) -\frac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{BC} \right)=\frac{1}{2}\left(  \overrightarrow{AD}-\overrightarrow{BC}  \right). 

Kuid          on kolineaarne vektoriga        , seega
                    . Siis

\overrightarrow{BC}

\overrightarrow{AC}

λ\overrightarrow{AD}=\overrightarrow{BC}

\overrightarrow{MN}=\frac{1}{2}\left( \overrightarrow{AD} -λ \overrightarrow{AD}\right) =\frac{1}{2}\left(  1-λ \right)\overrightarrow{AD}=k\overrightarrow{AD}

Seega MN ∥ AD.

ON

OFF

n = 5 (slider from 1 to 5, step 1)
A = (-3, 0)
D = (3, 0)
B = (-1.02, 1.92)
C = (0.98, 1.92)
trapets, a, b, d, c = Polygon(A, D, C, B)
f = Segment(B, D)
g = Segment(A, C)
M = Midpoint(A, C)
N = Midpoint(B, D)
h = Segment(A, N)
i = Segment(A, M)
j = Segment(M, N)