The midsegment of a triangle and of a trapezoid

The proofs of this course rest on definitions and on theorems already proved. From grades 8 and 9 you have:

• congruent triangles – SAS, ASA, SSS, and two sides with the angle opposite the larger of them (lesson 11 goes into it);
• the isosceles triangle – the base angles are equal; the bisector of the apex angle, the median to the base and the altitude to the base are one segment;
• angles – vertical angles are equal; corresponding and alternate angles between parallel lines are equal, and equal ones make the lines parallel; the angles of a triangle add up to $\displaystyle 180^{\circ}$, and an exterior angle is the sum of the two interior angles not next to it;
• the quadrilaterals – the parallelogram (opposite sides equal; the diagonals bisect each other; a quadrilateral with a pair of opposite sides parallel and equal is a parallelogram), the rectangle, the rhombus, the square, the kite and the trapezoid;
• the right-angled triangle – Pythagoras’ theorem; the median to the hypotenuse is half of it; the leg opposite $\displaystyle 30^{\circ}$ is half the hypotenuse.

A proof is a chain of claims, each with its reason:

• given – what the question says; to prove – the claim at the end of the chain;
• each row – a claim, and its reason: “given”, a definition, a theorem, or rows above it by their numbers;
• a figure helps to find the proof, but it proves nothing: what looks equal in it may not be.

A theorem says “if A, then B”. Its converse says “if B, then A” – a claim of its own, which needs its own proof and is not always true.

A midsegment of a triangle is a segment joining the midpoints of two sides.

Theorem: a midsegment of a triangle is parallel to the third side and equal to half of it.

Given: $\displaystyle D$ is the midpoint of $\displaystyle AB$, and $\displaystyle E$ the midpoint of $\displaystyle AC$.
To prove: $\displaystyle DE\parallel BC$ and $\displaystyle {DE}={\frac{1}{2}BC}$.

The proof below extends $\displaystyle DE$ beyond $\displaystyle E$ to a point $\displaystyle F$ with $\displaystyle {EF}={DE}$. Congruent triangles then show that $\displaystyle DBCF$ is a parallelogram, whose side $\displaystyle DF$ is parallel and equal to $\displaystyle BC$ – and $\displaystyle DE$ is half of $\displaystyle DF$.

Two theorems recognise a midsegment:

• a line through the midpoint of one side of a triangle, parallel to a second side, bisects the third side;
• a segment whose ends are on two sides of a triangle, parallel to the third side and equal to half of it, is a midsegment.

Why the first holds: the line through $\displaystyle D$, the midpoint of $\displaystyle AB$, parallel to $\displaystyle BC$ meets $\displaystyle AC$ at $\displaystyle E$. The midsegment from $\displaystyle D$ to the midpoint of $\displaystyle AC$ is parallel to $\displaystyle BC$ too – and through $\displaystyle D$ there is only one line parallel to $\displaystyle BC$. So the two are one line, and $\displaystyle E$ is the midpoint of $\displaystyle AC$.

In the triangle $\displaystyle ABC$, $\displaystyle D$ is the midpoint of $\displaystyle AB$ and $\displaystyle E$ the midpoint of $\displaystyle AC$.
$\displaystyle F$ is the midpoint of $\displaystyle BC$.
Prove: $\displaystyle ADFE$ is a parallelogram.

The sides of a triangle are $\displaystyle 10$, $\displaystyle 14$ and $\displaystyle 18$.
Its three midsegments make a small triangle.
Find the perimeter of the small triangle.

The midsegment of a trapezoid joins the midpoints of its two legs.

Theorem: the midsegment of a trapezoid is parallel to the bases and equal to half their sum.

Given: $\displaystyle ABCD$ is a trapezoid, $\displaystyle BC\parallel AD$; $\displaystyle E$ is the midpoint of $\displaystyle AB$, and $\displaystyle F$ the midpoint of $\displaystyle CD$.
To prove: $\displaystyle EF\parallel AD$ and $\displaystyle {EF}={\frac{1}{2}(AD+BC)}$.

The proof below draws $\displaystyle BF$ on until it meets the line $\displaystyle AD$ at $\displaystyle G$. Congruent triangles show that $\displaystyle F$ is the midpoint of $\displaystyle BG$ and $\displaystyle {DG}={BC}$, so $\displaystyle EF$ is a midsegment of the triangle $\displaystyle ABG$.

Theorem: in a trapezoid, a line through the midpoint of one leg, parallel to the bases, bisects the other leg.

The reason is the one of the triangle’s converse: the midsegment from that midpoint is parallel to the bases, and through a point there is only one line parallel to a given line.

Find the missing lengths.

$\displaystyle ABCD$ is any quadrilateral.
$\displaystyle E$, $\displaystyle F$, $\displaystyle G$ and $\displaystyle H$ are the midpoints of $\displaystyle AB$, $\displaystyle BC$, $\displaystyle CD$ and $\displaystyle DA$.
Prove: $\displaystyle EFGH$ is a parallelogram.
(A hint: draw the diagonal $\displaystyle AC$.)

$\displaystyle ABCD$ is a trapezoid, $\displaystyle BC\parallel AD$, with $\displaystyle {BC}={6}$ and $\displaystyle {AD}={14}$.
Its midsegment $\displaystyle EF$ meets the diagonal $\displaystyle AC$ at $\displaystyle K$ and the diagonal $\displaystyle BD$ at $\displaystyle L$.