Mid-Point and Its Converse

• By the Mid-Point Theorem:
The line segment joining the mid-points of two sides of a triangle is parallel to the third side and equal to half of its length.

$$DE = \frac{1}{2}BC$$

• Substitute $BC = 11.4\text{ cm}$:

$$DE = \frac{11.4}{2} = \mathbf{5.7\text{ cm}}$$

• Theorem Used: The Mid-Point Theorem.

• Let $D, E, F$ be the mid-points of $BC, CA, AB$ respectively in $\triangle ABC$.
• By the Mid-Point Theorem:
- $FE = \frac{1}{2}BC = BD = DC$
- $DE = \frac{1}{2}AB = AF = FB$
- $DF = \frac{1}{2}AC = AE = EC$
• Now examine the four triangles: $\triangle AFE, \triangle FBD, \triangle EDC, \triangle DEF$:
1. In $\triangle AFE$: sides are $AF, AE, FE$.
2. In $\triangle FBD$: sides are $FB, BD, FD$.
3. In $\triangle EDC$: sides are $ED, DC, EC$.
4. In $\triangle DEF$: sides are $DE, EF, FD$.
• In each triangle, the three sides correspond exactly to $\frac{1}{2}AB, \frac{1}{2}BC, \frac{1}{2}CA$.
• Therefore, by the SSS Congruence Criterion, all four triangles are congruent to one another: $\triangle AFE \cong \triangle FBD \cong \triangle EDC \cong \triangle DEF$. $\blacksquare$

• Let $ABCD$ be a rectangle. Diagonals of a rectangle are equal: $AC = BD$.
• Let $P, Q, R, S$ be the mid-points of sides $AB, BC, CD, DA$.
• By the Mid-Point Theorem:
- In $\triangle ABC$: $PQ = \frac{1}{2}AC$
- In $\triangle ADC$: $SR = \frac{1}{2}AC$
- In $\triangle ABD$: $SP = \frac{1}{2}BD$
- In $\triangle BCD$: $QR = \frac{1}{2}BD$
• Since $AC = BD$ (diagonals of a rectangle are equal):

$$PQ = SR = SP = QR = \frac{1}{2}AC$$

• A quadrilateral having all four sides equal is a rhombus.
• Therefore, $PQRS$ is a rhombus. $\blacksquare$

• Let $ABCD$ be a rhombus. The diagonals of a rhombus intersect at right angles: $AC \perp BD$.
• Let $P, Q, R, S$ be the mid-points of sides $AB, BC, CD, DA$.
• By Varignon's Theorem, $PQRS$ is a parallelogram with $PQ \parallel AC$ and $SP \parallel BD$.
• Since $AC \perp BD$, the angle between $AC$ and $BD$ is $90^\circ$.
• Because $PQ \parallel AC$ and $SP \parallel BD$, the angle between the intersecting sides $PQ$ and $SP$ must also be equal to the angle between $AC$ and $BD$, which is $90^\circ$:

$$\angle SPQ = 90^\circ$$

• A parallelogram having one angle equal to $90^\circ$ is a rectangle.
• Therefore, $PQRS$ is a rectangle. $\blacksquare$

• Construction: Draw $DG \parallel BF$ meeting $AC$ at $G$.
• In $\triangle ADG$:
$E$ is the mid-point of $AD$ (given), and $EF \parallel DG$ (by construction).
By the Converse of the Mid-Point Theorem, $F$ is the mid-point of $AG$:

$$AF = FG \quad \text{--- (1)}$$

• In $\triangle CBF$:
$D$ is the mid-point of $BC$ (since $AD$ is a median), and $DG \parallel BF$ (by construction).
By the Converse of the Mid-Point Theorem, $G$ is the mid-point of $FC$:

$$FG = GC \quad \text{--- (2)}$$

• Combining (1) and (2):

$$AF = FG = GC$$

• Therefore, side $AC$ is partitioned into three equal segments:

$$AC = AF + FG + GC = 3AF \implies \mathbf{AF = \frac{1}{3}AC} \quad \blacksquare$$

• Statement: If three or more parallel lines make equal intercepts on one transversal, they make equal intercepts on any other transversal.
• Application: To divide a given segment $AB$ into $n$ equal parts, draw an acute ray $AX$. On $AX$, mark $n$ equal segments $A_1, A_2, ..., A_n$ using a compass. Join $A_n$ to $B$. Through each point $A_k$, draw lines parallel to $A_nB$. By the Equal Intercept Theorem, these parallel lines make equal intercepts on $AB$, dividing it into $n$ equal parts.

• Join diagonal $AC$, meeting line segment $EF$ at point $G$.
• In $\triangle ADC$:
$E$ is the mid-point of $AD$, and $EG \parallel DC$ (since $AB \parallel DC$ and $EF$ is parallel to the bases).
By the Converse of Mid-Point Theorem, $G$ is the mid-point of $AC$.
By the Mid-Point Theorem:

$$EG = \frac{1}{2}DC \quad \text{--- (1)}$$

• In $\triangle ABC$:
$G$ is the mid-point of $AC$, and $F$ is the mid-point of $BC$.
By the Mid-Point Theorem:

$$GF = \frac{1}{2}AB \quad \text{--- (2)}$$

• Adding (1) and (2):

$$EF = EG + GF = \frac{1}{2}DC + \frac{1}{2}AB = \mathbf{\frac{1}{2}(AB + DC)} \quad \blacksquare$$