Mid-Point Theorem

Consider $\triangle AEF$ and $\triangle CDF$.

Therefore, by the SAS (Side-Angle-Side) congruence criterion:

Since the triangles are congruent, their corresponding parts are equal by CPCT (Corresponding Parts of Congruent Triangles).

We are given that E is the mid-point of $\overline{AB}$.

From equation (1) and the fact that $AE = EB$, we have:

Now consider equation (2), $\angle EAF = \angle DCF$. These angles can be viewed as alternate interior angles formed by the transversal $\overline{AC}$ intersecting lines $\overline{AB}$ (containing segment $\overline{AE}$) and $\overline{DC}$.

Since the alternate interior angles are equal, the lines must be parallel:

Since $\overline{EB}$ is a part of the line $\overline{AB}$, we can say:

Now, consider the quadrilateral EBCD. We have shown that:

• One pair of opposite sides, $\overline{EB}$ and $\overline{CD}$, are equal ($EB = CD$).
• The same pair of opposite sides, $\overline{EB}$ and $\overline{DC}$, are parallel ($EB \parallel DC$).

A quadrilateral in which one pair of opposite sides is both equal and parallel is a parallelogram.

Therefore, EBCD is a parallelogram.

Since EBCD is a parallelogram, its other pair of opposite sides must also be equal and parallel.

Thus, $\overline{ED} \parallel \overline{BC}$ and $ED = BC$.

From $ED \parallel BC$, and knowing that $\overline{EF}$ is a part of the line segment $\overline{ED}$, we can conclude that:

Also, $ED = BC$. From our construction, the line segment $\overline{ED}$ is formed by extending $\overline{EF}$ such that $EF = FD$. Thus, $ED = EF + FD$.

Since $EF = FD$, we have $ED = EF + EF = 2EF$.

Substituting this into the equation $ED = BC$:

Dividing both sides by 2, we get:

Hence, the Mid-Point Theorem is proved: the line segment joining the mid-points of two sides of a triangle is parallel to the third side and is half the length of the third side.

Consider the quadrilateral EBCD. By construction, we have $CD \parallel AB$. Since E is a point on $\overline{AB}$, we can say $CD \parallel EB$.

We are given that the line through E is parallel to $\overline{BC}$. Since F lies on ED (part of the extended line through E parallel to BC), we can say $ED \parallel BC$.

So, in quadrilateral EBCD, both pairs of opposite sides are parallel: $\overline{CD} \parallel \overline{EB}$ and $\overline{ED} \parallel \overline{BC}$. Therefore, EBCD is a parallelogram.

In a parallelogram, opposite sides are equal in length. Thus, the opposite sides $\overline{CD}$ and $\overline{EB}$ must be equal:

We are given that E is the mid-point of side $\overline{AB}$. This means:

From equation (1) and the fact that $AE = EB$, we can equate AE and CD:

Now, let's consider $\triangle AEF$ and $\triangle CDF$.

Since $AB \parallel CD$ (as $EB$ is part of $AB$ and $EB \parallel CD$) and $\overline{AC}$ is a transversal, the alternate interior angles formed are equal:

(Note: $\angle EAF$ is $\angle BAC$, and $\angle DCF$ is $\angle ACD$)

The angles $\angle AFE$ and $\angle CFD$ are formed by the intersection of lines $\overline{AD}$ and $\overline{AC}$. These are vertically opposite angles.

We have already shown that:

From (2), (3), and (4), by the AAS (Angle-Angle-Side) congruence criterion (Angle $\angle EAF$, Angle $\angle AFE$, and side AE), we can conclude that:

Since the triangles are congruent, their corresponding parts are equal by CPCT:

This equality $AF = CF$ means that point F divides the side $\overline{AC}$ into two equal segments. Therefore, F is the mid-point of the side $\overline{AC}$.

Hence, the converse of the Mid-Point Theorem is proved.

Given:

A quadrilateral ABCD.

P, Q, R, and S are the mid-points of sides $\overline{AB}$, $\overline{BC}$, $\overline{CD}$, and $\overline{DA}$ respectively.

To Prove:

PQRS is a parallelogram.

Construction:

Join the diagonal $\overline{AC}$.

Proof:

Consider $\triangle ABC$.

P is the mid-point of $\overline{AB}$ and Q is the mid-point of $\overline{BC}$.

By the Mid-Point Theorem applied to $\triangle ABC$:

Now consider $\triangle ADC$.

S is the mid-point of $\overline{DA}$ and R is the mid-point of $\overline{DC}$.

By the Mid-Point Theorem applied to $\triangle ADC$:

From statement (i) and statement (ii):

• Since $\overline{PQ} \parallel \overline{AC}$ and $\overline{SR} \parallel \overline{AC}$, lines parallel to the same line are parallel to each other. Thus, $\overline{PQ} \parallel \overline{SR}$.
• Since $PQ = \frac{1}{2} AC$ and $SR = \frac{1}{2} AC$, by substitution, $PQ = SR$.

Now, consider the quadrilateral PQRS. We have shown that one pair of opposite sides, $\overline{PQ}$ and $\overline{SR}$, are both equal in length ($PQ = SR$) and parallel ($\overline{PQ} \parallel \overline{SR}$).

A quadrilateral in which one pair of opposite sides is both equal and parallel is a parallelogram (Converse of a property of parallelogram).

Therefore, PQRS is a parallelogram.

Further results:

We could have similarly joined the diagonal $\overline{BD}$ and applied the Mid-Point Theorem to $\triangle ABD$ and $\triangle CBD$ to prove that $\overline{PS} \parallel \overline{BD}$ and $PS = \frac{1}{2} BD$, and $\overline{QR} \parallel \overline{BD}$ and $QR = \frac{1}{2} BD$. This would lead to $PS \parallel QR$ and $PS = QR$, which also proves that PQRS is a parallelogram.

The specific type of parallelogram PQRS depends on the original quadrilateral ABCD:

• If ABCD is a rectangle, PQRS is a rhombus.
• If ABCD is a rhombus, PQRS is a rectangle.
• If ABCD is a square, PQRS is a square.
• If ABCD is an isosceles trapezium, PQRS is a rhombus.

Given:

In $\triangle ABC$, D, E, and F are the mid-points of $\overline{AB}$, $\overline{BC}$, and $\overline{CA}$ respectively.

$AB = 7$ cm, $BC = 8$ cm, $CA = 9$ cm.

To Find:

The perimeter of $\triangle DEF$.

Solution:

We can apply the Mid-Point Theorem to relate the sides of $\triangle DEF$ to the sides of $\triangle ABC$.

• Since D and E are the mid-points of $\overline{AB}$ and $\overline{BC}$ respectively, by the Mid-Point Theorem, $\overline{DE} \parallel \overline{AC}$ and $DE = \frac{1}{2} AC$.
• Since E and F are the mid-points of $\overline{BC}$ and $\overline{CA}$ respectively, by the Mid-Point Theorem, $\overline{EF} \parallel \overline{AB}$ and $EF = \frac{1}{2} AB$.
• Since F and D are the mid-points of $\overline{CA}$ and $\overline{AB}$ respectively, by the Mid-Point Theorem, $\overline{FD} \parallel \overline{BC}$ and $FD = \frac{1}{2} BC$.

Now, we substitute the given side lengths of $\triangle ABC$ to find the lengths of the sides of $\triangle DEF$:

The perimeter of $\triangle DEF$ is the sum of the lengths of its sides:

Thus, the perimeter of $\triangle DEF$ is 12 cm.

Alternate Method:

We can also find the perimeter of $\triangle ABC$ first:

From the application of the Mid-Point Theorem, we found that $DE = \frac{1}{2} AC$, $EF = \frac{1}{2} AB$, and $FD = \frac{1}{2} BC$.

So, Perimeter($\triangle DEF$) = $DE + EF + FD = \frac{1}{2} AC + \frac{1}{2} AB + \frac{1}{2} BC = \frac{1}{2} (AC + AB + BC)$

Perimeter($\triangle DEF$) = $\frac{1}{2} \times$ Perimeter($\triangle ABC$).

Both methods yield the same result.