The mid-point theorem

Mark the mid­points of two sides of any tri­an­gle and join them. How­ever the tri­an­gle is shaped, the join­ing seg­ment always runs par­al­lel to the third side and is exactly half as long. This fact, the mid-point the­o­rem, is one of the most used results in school geom­e­try. In this les­son we prove it with con­gru­ent tri­an­gles and a par­al­lel­o­gram, prove its con­verse, and use both to solve prob­lems on tri­an­gles, trapez­i­ums and quadri­lat­er­als.

The mid-point the­o­rem#

The­o­rem. The seg­ment join­ing the mid­points of two sides of a tri­an­gle is par­al­lel to the third side and equal to half of it.

Proof. In tri­an­gle $ABC$, let $D$ and $E$ be the mid­points of $AB$ and $AC$. We must show $DE \parallel BC$ and $\displaystyle DE = \frac{1}{2}BC$.

Through $C$ draw a line par­al­lel to $BA$, and extend $DE$ to meet it at $F$.

In $\triangle AED$ and $\triangle CEF$:

• $AE = CE$ ($E$ is the mid­point of $AC$)
• $\angle AED = \angle CEF$ (ver­ti­cally oppo­site angles)
• $\angle DAE = \angle FCE$ (alter­nate angles, since $CF \parallel BA$ and $AC$ is a trans­ver­sal)

So $\triangle AED \cong \triangle CEF$ by ASA. By CPCT, $AD = CF$ and $DE = EF$.

Now $AD = BD$ ($D$ is the mid­point), so $BD = CF$. Also $BD \parallel CF$. A quadri­lat­eral with one pair of oppo­site sides equal and par­al­lel is a par­al­lel­o­gram, so $BCFD$ is a par­al­lel­o­gram. There­fore $DF \parallel BC$ and $DF = BC$.

Since $DE = EF$, $E$ is the mid­point of $DF$, so $\displaystyle DE = \frac{1}{2}DF = \frac{1}{2}BC$, and $DE$ lies along $DF$, which is par­al­lel to $BC$.

Exam­ple 1#

In tri­an­gle $ABC$, $D$ and $E$ are the mid­points of $AB$ and $AC$, and $BC = 14$ cm. Find $DE$.

By the mid-point the­o­rem, $\displaystyle DE = \frac{1}{2} \times 14 = 7$ cm.

Exam­ple 2#

The sides of a tri­an­gle are $8$ cm, $10$ cm and $12$ cm. Find the perime­ter of the tri­an­gle formed by join­ing the mid­points of its sides.

Each side of the new tri­an­gle joins two mid­points, so it is half of the side it is par­al­lel to. The new sides are $4$, $5$ and $6$ cm, and the perime­ter is $4 + 5 + 6 = 15$ cm, half the orig­i­nal perime­ter of $30$ cm.

The three seg­ments also cut the tri­an­gle into four smaller tri­an­gles, all con­gru­ent to each other (each has sides $4$, $5$ and $6$ cm).

The con­verse#

Con­verse. The line drawn through the mid­point of one side of a tri­an­gle, par­al­lel to a sec­ond side, bisects the third side.

So in tri­an­gle $ABC$, if $D$ is the mid­point of $AB$ and the line through $D$ par­al­lel to $BC$ meets $AC$ at $E$, then $E$ is the mid­point of $AC$. (If it were not, the seg­ment from $D$ to the true mid­point would also be par­al­lel to $BC$, giv­ing two dif­fer­ent lines through $D$ par­al­lel to $BC$, which is impos­si­ble.)

Exam­ple 3#

$ABCD$ is a trapez­ium with $AB \parallel DC$, $AB = 16$ cm and $DC = 10$ cm. $E$ is the mid­point of $AD$, and the line through $E$ par­al­lel to $AB$ meets $BC$ at $F$. Show that $F$ is the mid­point of $BC$ and find $EF$.

Join $AC$, cross­ing $EF$ at $G$.

In tri­an­gle $ADC$: $E$ is the mid­point of $AD$ and $EG \parallel DC$, so by the con­verse $G$ is the mid­point of $AC$, and by the the­o­rem $\displaystyle EG = \frac{1}{2}DC = 5$ cm.

In tri­an­gle $CAB$: $G$ is the mid­point of $CA$ and $GF \parallel AB$, so $F$ is the mid­point of $CB$, and $\displaystyle GF = \frac{1}{2}AB = 8$ cm.

So $F$ is the mid­point of $BC$, and

$$EF = EG + GF = 5 + 8 = 13 \text{ cm}$$

In gen­eral, the seg­ment join­ing the mid­points of the non-par­al­lel sides of a trapez­ium is par­al­lel to the par­al­lel sides and equal to half their sum.

Mid­points of a quadri­lat­eral#

Exam­ple 4#

$P$, $Q$, $R$, $S$ are the mid­points of the sides $AB$, $BC$, $CD$, $DA$ of any quadri­lat­eral $ABCD$. Prove that $PQRS$ is a par­al­lel­o­gram.

Draw the diag­o­nal $AC$.

In tri­an­gle $ABC$, $P$ and $Q$ are mid­points of $AB$ and $BC$, so $PQ \parallel AC$ and $\displaystyle PQ = \frac{1}{2}AC$.

In tri­an­gle $ADC$, $S$ and $R$ are mid­points of $AD$ and $DC$, so $SR \parallel AC$ and $\displaystyle SR = \frac{1}{2}AC$.

Hence $PQ \parallel SR$ and $PQ = SR$. One pair of oppo­site sides of $PQRS$ is equal and par­al­lel, so $PQRS$ is a par­al­lel­o­gram.

The sides of $PQRS$ are half the diag­o­nals of $ABCD$. So if the diag­o­nals are $10$ cm and $14$ cm, the par­al­lel­o­gram has sides $5$ cm and $7$ cm and perime­ter $24$ cm: the perime­ter equals the sum of the diag­o­nals.

Prac­tice#

1. In tri­an­gle $PQR$, $M$ and $N$ are the mid­points of $PQ$ and $PR$, and $MN = 6.5$ cm. Find $QR$.
2. The sides of a tri­an­gle are $7$ cm, $9$ cm and $12$ cm. Find the perime­ter of the tri­an­gle formed by join­ing the mid­points of its sides.
3. $D$, $E$, $F$ are the mid­points of the sides of a tri­an­gle whose perime­ter is $36$ cm. Find the perime­ter of tri­an­gle $DEF$.
4. In tri­an­gle $ABC$, $D$ is the mid­point of $AB$, and the line through $D$ par­al­lel to $BC$ meets $AC$ at $E$. If $AC = 11$ cm, find $AE$.
5. The par­al­lel sides of a trapez­ium are $12$ cm and $20$ cm. Find the length of the seg­ment join­ing the mid­points of the other two sides.
6. In a trapez­ium, the seg­ment join­ing the mid­points of the non-par­al­lel sides is $15$ cm, and one par­al­lel side is $18$ cm. Find the other par­al­lel side.
7. The diag­o­nals of a quadri­lat­eral are $9$ cm and $13$ cm. Find the perime­ter of the fig­ure formed by join­ing the mid­points of its sides.
8. $ABCD$ is a rec­tan­gle and $P$, $Q$, $R$, $S$ are the mid­points of $AB$, $BC$, $CD$, $DA$. Prove that $PQRS$ is a rhom­bus.

Answers#