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Prove that the line joining the midpoint of parallel sides of a trapezium passes through the point of intersection of diagonals.

I want to use theorems in geometry to solve this question.

The method using vectors is given here.

Let $ABCD$ be the trapezium and let $O$ be the point of intersection of diagonals. I need to prove that one of the lines through $O$ can pass through both midpoints of adjacent sides. Let $AB$ and $CD$ be the parallel lines.

I started by considering a line through $O$ that passes through midpoint $E$ of $AB$. I need to show that if the line is extended, it also passes through midpoint $F$ of $CD$.

I tried using similarity of triangles but it did not help much.

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1 Answer 1

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1) Let the line $MO$ intersects the sides of the trapezoid at the points $E$ and $F$. Then $$\triangle BME \sim \triangle AMF \Rightarrow\frac {BE}{AF}=\frac {ME}{MF}$$

2) $$\triangle EMC \sim \triangle FMD \Rightarrow\frac {ME}{MF}=\frac {EC}{FD}\Rightarrow\frac {BE}{AF}=\frac {EC}{FD}$$

3) $$\triangle AOF \sim \triangle COE \Rightarrow\frac {EC}{AF}=\frac {EO}{OF}$$

$$\triangle BEO \sim \triangle DFO \Rightarrow\frac {BE}{FD}=\frac {EO}{OF} \Rightarrow\frac {BE}{FD}=\frac {EC}{AF}$$ So $$\frac {BE}{AF}=\frac {EC}{FD}$$ and $$\frac{BE}{FD}=\frac {EC}{AF}$$

Then the four points $M,E,O,F$ lie on a straight line

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  • $\begingroup$ To whom is this theorem attributed? $\endgroup$
    – user366533
    Mar 3, 2017 at 20:14
  • $\begingroup$ After BE/AF=EC/FD and BE/FD=EC/AF, how did we get that the four points are on a straight line ? $\endgroup$
    – Fin27
    Jun 1, 2023 at 6:11

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