Triangles - Apply converse of Basic Proportionality Theorem to establish parallelism

In $\triangle ABC$, $D$ and $E$ are points on the sides $AB$ and $AC$ respectively. If $AD = 5.7\text{ cm}$, $DB = 9.5\text{ cm}$, $AE = 4.8\text{ cm}$, and $EC = 8.0\text{ cm}$, determine whether $DE \parallel BC$.

To check if $DE \parallel BC$, we must calculate the ratios of the segments on both sides:

1. Calculate the ratio on side $AB$: $\frac{AD}{DB} = \frac{5.7}{9.5} = \frac{57}{95}$ Dividing both by $19$: $\frac{57 \div 19}{95 \div 19} = \frac{3}{5}$

2. Calculate the ratio on side $AC$: $\frac{AE}{EC} = \frac{4.8}{8.0} = \frac{48}{80}$ Dividing both by $16$: $\frac{48 \div 16}{80 \div 16} = \frac{3}{5}$

Since $\frac{AD}{DB} = \frac{AE}{EC} = \frac{3}{5}$, by the Converse of Basic Proportionality Theorem, $DE \parallel BC$.

We compare the ratio of the parts created by the line $DE$ on side $AB$ and side $AC$. Because both ratios simplify to the same value ($0.6$ or $3/5$), the line $DE$ divides the sides proportionally, satisfying the condition for parallelism.

In $\triangle PQR$, $S$ and $T$ are points on $PQ$ and $PR$ respectively such that $PQ = 1.28\text{ cm}$, $PR = 2.56\text{ cm}$, $PS = 0.18\text{ cm}$, and $PT = 0.36\text{ cm}$. Show that $ST \parallel QR$.

We are given the total lengths of the sides and the lengths of the upper segments. We can use the alternative form of the Converse of BPT ratio:

1. Ratio on side $PQ$: $\frac{PS}{PQ} = \frac{0.18}{1.28} = \frac{18}{128} = \frac{9}{64}$

2. Ratio on side $PR$: $\frac{PT}{PR} = \frac{0.36}{2.56} = \frac{36}{256} = \frac{9}{64}$

Since $\frac{PS}{PQ} = \frac{PT}{PR}$, the segments are divided in the same ratio. Therefore, by the Converse of Basic Proportionality Theorem, $ST \parallel QR$.

The Converse of BPT can be applied using the ratio of the small segment to the whole side length. If $\frac{\text{Part}}{\text{Whole}}$ is consistent for both sides, the line joining the points is parallel to the base.

Given: $LX = 2\text{ cm}$, $XM = 6\text{ cm}$ $LY = 3\text{ cm}$, $YN = 9\text{ cm}$

Step 1: Calculate the ratio on side $LM$: $\frac{LX}{XM} = \frac{2}{6} = \frac{1}{3}$

Step 2: Calculate the ratio on side $LN$: $\frac{LY}{YN} = \frac{3}{9} = \frac{1}{3}$

Step 3: Compare ratios: Since $\frac{LX}{XM} = \frac{LY}{YN} = \frac{1}{3}$, by the Converse of Basic Proportionality Theorem, $XY \parallel MN$.

By checking the ratios of the segments created by the line $XY$ on the two sides of the triangle, we find they are identical ($1:3$). This satisfies the condition for the converse of BPT.

Given: $AD = 3\text{ cm}$, $AB = 12\text{ cm}$ $AE = 4\text{ cm}$, $AC = 16\text{ cm}$

Step 1: Calculate the ratio $\frac{AD}{AB}$: $\frac{AD}{AB} = \frac{3}{12} = \frac{1}{4}$

Step 2: Calculate the ratio $\frac{AE}{AC}$: $\frac{AE}{AC} = \frac{4}{16} = \frac{1}{4}$

Step 3: Conclusion: Since $\frac{AD}{AB} = \frac{AE}{AC}$, the line $DE$ divides the sides $AB$ and $AC$ in the same ratio. Therefore, $DE \parallel BC$ by the Converse of Basic Proportionality Theorem.

Even when full side lengths are given, the ratio of the small part to the whole side must be equal for both sides to prove parallelism.