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Home » Quadrangles » Trapezoids » Lines that Intersect on a Trapezoid's Midsegment

Lines that Intersect on a Trapezoid's Midsegment

Last updated: Feb 8, 2020 by Ido Sarig · This website generates income via ads and uses cookies · Terms of use · Privacy policy

A reader wanted some help and sent me the following problem, involving lines that intersect on a trapezoid's midsegment:

Problem:

EF is the midsegment of trapezoid ABCD, and the two line segments KM and LN intersect at point O, which lies on EF. Show that |KO|=|OM| and |NO|=|OL| .

Lines that Intersect on a Trapezoid's Midsegment

The original problem is quite easy to prove, using the converse trapezoid midsegment theorem, and we will prove a couple of additional things- |KN| =|LM|, and |KL|=|NM|.

Strategy

The problem statement begins with saying the EF is a midsegment in a trapezoid, so we will use the properties of such midsegments. As we have proven, the midsegment of a trapezoid is parallel to both of the bases. So EF is parallel to DC. Now let's look at the quadrilateral AKMD. It is also a trapezoid, since AK is just a sub-segment of AB and DM is a sub-segment of DC. As ABCD is a trapezoid, AB||CD, and so AK||DM, and since EF|| DC, it also follows that EF||DM.

Now we will use the converse trapezoid midsegment theorem , and apply it to trapezoid AKMD. The converse trapezoid midsegment theorem states that in a trapezoid, a line that intersects the midpoint of one of the legs and is parallel to one of the bases, it is a midsegment, and it intersects the other leg's midpoint. So in trapezoid AKMD, EF intersects the midpoint of leg AD, and is parallel to DM, making it a midsegment, and the result is that it intersects the other leg (KM) at its midpoint, proving that |KO|=|OM|.

We can repeat this entire process for the quadrilateral LBCN. LB is just a sub-segment of AB and CN is a sub-segment of DC. As ABCD is a trapezoid, AB||CD, and so LB||CN, and since EF|| DC, it also follows that EF||CN. Applying the converse trapezoid midsegment theorem to LBCN, EF intersects the midpoint of leg BC, and is parallel to CN, making it a midsegment, and the result is that it intersects the other leg (LN) at its midpoint, proving that |NO|=|OL|.

Additional proofs

So we have proven what was required in the original problem, now let's prove the additional things.

Consider triangles ΔOKN and ΔOML. We have just shown that |KO|=|OM| & |NO|=|OL|. ∠KON and ∠MOL are vertical angles, so they are congruent, and so ΔOKN ≅ ΔOML by the Side-Angle-Side postulate, and |KN| =|LM| as corresponding sides in congruent triangles.

Similarly, if we look at triangles ΔOKL and ΔOMN - we have just shown that |KO|=|OM| & |NO|=|OL|. ∠KOL and ∠MON are vertical angles, so they are congruent, and so ΔOKL≅ ΔOMN by the Side-Angle-Side postulate, and |KL|=|NM| as corresponding sides in congruent triangles.

This is an interesting result - in a trapezoid, any two lines that intersect at a point that lies on the trapezoid's midsegment, cut congruent segments from the trapezoid's bases.

Proof

(1) EF ||DC //Given, EF is the midsegment of trapezoid ABCD
(2) |AE|=|ED| //Given, EF is the midsegment of trapezoid ABCD
(3) AK||DM //Given, ABCD is a trapezoid, AK and DM are sub-segments of parallel lines
(4) |KO|=|OM| //Converse midsegment theorem in trapezoid AKMD

(5) |BF|=|FC| //Given, EF is the midsegment of trapezoid ABCD
(6) LB||CN //Given, ABCD is a trapezoid, LB and CN are sub-segments of parallel lines
(7) |NO|=|OL| //Converse midsegment theorem in trapezoid LBCN

(8) ∠KON ≅ ∠MOL //Vertical angles
(9) ΔOKN ≅ ΔOM //(4), (7), (8), Side-Angle-Side postulate
(10) |KN| =|LM| //Corresponding sides in congruent triangles (CPCTC)
(11)∠KOL ≅ ∠MON //Vertical angles
(12) ΔOKL≅ ΔOMN //(4), (7), (10), Side-Angle-Side postulate
(13) |KL|=|NM| //Corresponding sides in congruent triangles (CPCTC)

« Converse of the Trapezoid Midsegment Theorem
Angles of Intersecting Chords »

About the Author

Ido Sarig is a high-tech executive with a BSc degree in Computer Engineering. His goal is to help you develop a better way to approach and solve geometry problems. You can contact him at [email protected]

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Welcome to Geometry Help! I'm Ido Sarig, a high-tech executive with a BSc degree in Computer Engineering and an MBA degree in Management of Technology. I'm here to tell you that geometry doesn't have to be so hard! My goal with this website is to help you develop a better way to approach and solve geometry problems, even if spatial awareness is not your strongest quality. Read More…

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