4-gons (Quadrilaterals) - Apply midpoint theorem and its converse in geometric proof problems

In $\triangle ABC$, $D, E$ and $F$ are the mid-points of sides $AB, BC$ and $CA$ respectively. If $AB = 6 \text{ cm}$, $BC = 7 \text{ cm}$ and $AC = 8 \text{ cm}$, find the perimeter of $\triangle DEF$.

1. By the Mid-point Theorem, $DE$ joins mid-points of $AB$ and $BC$, so $DE = \frac{1}{2} AC = \frac{1}{2} \times 8 = 4 \text{ cm}$.
2. Similarly, $EF$ joins mid-points of $BC$ and $CA$, so $EF = \frac{1}{2} AB = \frac{1}{2} \times 6 = 3 \text{ cm}$.
3. $DF$ joins mid-points of $AB$ and $CA$, so $DF = \frac{1}{2} BC = \frac{1}{2} \times 7 = 3.5 \text{ cm}$.
4. Perimeter of $\triangle DEF = DE + EF + DF = 4 + 3 + 3.5 = 10.5 \text{ cm}$.

This approach uses the Mid-point Theorem to determine each side of the inner triangle by taking half the length of the side of the original triangle to which it is parallel.

In $\triangle ABC$, $AD$ is the median and $E$ is the mid-point of $AD$. $BE$ is produced to meet $AC$ at $F$. Show that $AF = \frac{1}{3} AC$.

1. Draw a line $DG \parallel BF$ meeting $AC$ at $G$.
2. In $\triangle BCF$, $D$ is the mid-point of $BC$ (since $AD$ is a median) and $DG \parallel BF$. By the Converse of Mid-point Theorem, $G$ is the mid-point of $FC$. Thus, $FG = GC$.
3. In $\triangle ADG$, $E$ is the mid-point of $AD$ and $EF \parallel DG$ (as $EF$ is part of $BF$). By the Converse of Mid-point Theorem, $F$ is the mid-point of $AG$. Thus, $AF = FG$.
4. Combining the results: $AF = FG = GC$. Since $AF + FG + GC = AC$, we have $3 \times AF = AC$.
5. Therefore, $AF = \frac{1}{3} AC$.

This solution utilizes a construction to create a new triangle where the Converse of the Mid-point Theorem can be applied twice to show that the segments on side $AC$ are equal.

Let $BD$ intersect $EF$ at point $G$.

1. In $\triangle ABD$, $E$ is the mid-point of $AD$ (Given).
2. $EG \parallel AB$ (Since $EF \parallel AB$).
3. By the Converse of Midpoint Theorem, $G$ is the mid-point of $BD$.

Now, in $\triangle BCD$:

1. $G$ is the mid-point of $BD$ (Proved above).
2. $GF \parallel DC$ (Since $EF \parallel AB$ and $AB \parallel DC$, so $EF \parallel DC$).
3. By the Converse of Midpoint Theorem, $F$ is the mid-point of $BC$.

We use the converse of the midpoint theorem twice: first in the triangle involving the left side and diagonal to locate the diagonal's midpoint, then in the triangle involving the right side and diagonal to prove $F$ is the midpoint.

1. In $\triangle ABC$, $P$ and $Q$ are mid-points of $AB$ and $BC$. By Midpoint Theorem, $PQ \parallel AC$ and $PQ = \frac{1}{2} AC$.
2. Similarly, in $\triangle ADC$, $SR \parallel AC$ and $SR = \frac{1}{2} AC$. Thus, $PQ \parallel SR$ and $PQ = SR$, so $PQRS$ is a parallelogram.
3. In a rhombus, diagonals $AC$ and $BD$ intersect at $90^\circ$ (say at $O$).
4. Since $PQ \parallel AC$ and $PS \parallel BD$, the angle between $PQ$ and $PS$ will be the same as the angle between $AC$ and $BD$.
5. Therefore, $\angle P = 90^\circ$.
6. A parallelogram with one angle $90^\circ$ is a rectangle. Hence, $PQRS$ is a rectangle.

Joining midpoints of any quadrilateral forms a parallelogram. Since the diagonals of a rhombus are perpendicular, the sides of the midpoint parallelogram (which are parallel to the diagonals) must also be perpendicular, making it a rectangle.