Lesson One property of a trapezoid

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One property of a trapezoid

Problem 1

In a trapezoid, the straight line passing through the intersection point of the two lateral sides and the intersection point of its diagonals bisects each base of the trapezoid. Prove.

Proof

Let ABCD be a trapezoid (Figure 1); G be the
intersection point of its lateral sides AD and BC;
E and F be the midpoints of its bases AB and DC;
O be the intersection point of its diagonals AC and BD.

The statement is that the four points G, E, O and F
lie in one straight line.

We will prove the statement in two steps.

Figure 1. To the Problem 1

Figure 2a. To the Step 1

Figure 2b. To the Step 2

Step 1 (Figure 2a). The three points G, E and F lie in one straight line. In other words, for any trapezoid the intersection point of its lateral sides and the midpoints of its bases lie in one straight line.

It was proved in the lesson One property of a median in a triangle that the median of a triangle drawn from the vertex to the base bisects every segment parallel to the base and con-
necting two other sides. In a triangle ABG the straight segment GF is the median, while the segment DC is parallel to the base AB. Hence, the median GF intersects the base DC of the trapezoid at the midpoint of the segment DC. Thus the intersection points of the trapezoid's lateral sides and the midpoints of its two bases lie in one straight line. The Step 1 statement is proved.

Step 2 (Figure 2b). The three points E, O and F lie in one straight line. In other words, for any trapezoid the straight line passing through the midpoint of a base and the intersection point of its diagonals bisects the other base.

Indeed, let the straight line EF (Figure 2b) passes through the midpoint E of the base DC and the intersection point O of the diagonals. Then the triangles OEC and OFA are similar in accordance to the AA similarity test for triangles, because they have two pairs of congruent angles. Therefore, the corresponding sides of these triangles ae proportional:

abs%28OF%29%2Fabs%28OE%29

. (1)

Also, the triangles OED and OFB are similar by the same reason. Therefore, their corresponding sides are proportional too:

abs%28OF%29%2Fabs%28OE%29

. (2)

From proportions (1) and (2) we have

abs%28AF%29%2Fabs%28EC%29

. (3)

Since |EC| = |DE|, the proportion (3) implies |AF| = |FB|. It means that the point F is the midpoint of the segment AB.
Thus the midpoints of the bases of the trapezoid and the intersection point of its diagonals lie in one straight line. The Step 2 statement is proved.

From the proved statements of the Step 1 and the Step 2 we get the proof to the statement of the Problem 1.

Summary

For any trapezoid, the midpoints of its bases, the intersection point of its diagonals and the intersection point of its lateral sides lie in one straight line.

My other lessons on similar triangles in this site are
    - Similar triangles,
    - Similarity tests for triangles,
    - Proofs of Similarity tests for triangles,
    - In a triangle a straight line parallel to its side cuts off a similar triangle,
    - Problems on similar triangles,
    - Similarity tests for right-angled triangles,
    - Problems on similarity for right-angled triangles,
    - Problems on similarity for right-angled and acute triangles,
    - One property of a median in a triangle   and
    - Miscellaneous problems on similar triangles
under the current topic, and
    - Solved problems on similar triangles
under the topic Geometry of the section Word problems.

To navigate over all topics/lessons of the Online Geometry Textbook use this file/link GEOMETRY - YOUR ONLINE TEXTBOOK.

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