Q5:

In Fig. 6.20, $DE \parallel OQ$ and $DF \parallel OR$. Show that $EF \parallel QR$.

Given: In $\triangle PQR$, we have points $D, E,$ and $F$ on sides $PQ, PR,$ and $QR$ respectively (based on the standard configuration of this theorem). Specifically, $DE \parallel OQ$ and $DF \parallel OR$, where $O$ is a point inside the triangle.

To Prove: $EF \parallel QR$.

Step 1: Applying Thales Theorem (Basic Proportionality Theorem) in $\triangle POQ$

We are given that $DE \parallel OQ$. According to the Basic Proportionality Theorem (BPT), if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

In $\triangle POQ$, since $DE \parallel OQ$:

$\frac{PD}{DQ} = \frac{PE}{EO}$ --- (Equation 1) [By Basic Proportionality Theorem]

Step 2: Applying Thales Theorem in $\triangle POR$

We are given that $DF \parallel OR$. Applying the Basic Proportionality Theorem in $\triangle POR$:

$\frac{PD}{DQ} = \frac{PF}{FR}$ --- (Equation 2) [By Basic Proportionality Theorem]

Step 3: Comparing the Equations

From Equation 1 and Equation 2, we observe that the left-hand side (LHS) of both equations is identical ($\frac{PD}{DQ}$). Therefore, we can equate the right-hand sides (RHS):

$\frac{PE}{EO} = \frac{PF}{FR}$ --- (Equation 3) [By Euclid's Axiom: Things which are equal to the same thing are equal to one another]

Step 4: Applying the Converse of the Basic Proportionality Theorem

In $\triangle PQR$, we have established that $\frac{PE}{EO} = \frac{PF}{FR}$.

According to the Converse of the Basic Proportionality Theorem, if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Since $\frac{PE}{EO} = \frac{PF}{FR}$ in $\triangle PQR$, it follows that the line segment $EF$ must be parallel to the base $QR$.

Therefore, $EF \parallel QR$.

Final Answer: Since the ratios of the segments on sides $PQ$ and $PR$ are equal, by the converse of the Basic Proportionality Theorem, $EF \parallel QR$.