In ΔPQR, S and T are points on PQ and PR, respectively, such that ST ∥ QR and ST divides the ΔPQR into two parts of equal areas. Then the ratio of PS and QS is:

The correct answer option is $1:\sqrt{2}-1$.

Step-by-step Explanation:

Let us consider triangle $△\mathrm{PQR}$ where a line segment $\mathrm{ST}$ is drawn parallel to $\mathrm{QR}$, with $S$ on $\mathrm{PQ}$ and $T$ on $\mathrm{PR}$.

Step 1: Prove Similarity of Triangles

Since $\mathrm{ST}\parallel \mathrm{QR}$, we have:

$\angle \mathrm{PST}=\angle \mathrm{PQR}$ (Corresponding angles)

$\angle \mathrm{PTS}=\angle \mathrm{PRQ}$ (Corresponding angles)

$\angle P$ is common to both $△\mathrm{PST}$ and $△\mathrm{PQR}$.

Therefore, by the AA (Angle-Angle) similarity criterion:

$△\mathrm{PST}\sim △\mathrm{PQR}$

Step 2: Relation Between Areas and Side Ratios

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides:

$\frac{\text{Area}(△\mathrm{PST})}{\text{Area}(△\mathrm{PQR})}={\left(\frac{\mathrm{PS}}{\mathrm{PQ}}\right)}^2$

Step 3: Use the Given Condition of Areas

It is given that $\mathrm{ST}$ divides $△\mathrm{PQR}$ into two parts of equal area. This means the area of $△\mathrm{PST}$ is half of the area of $△\mathrm{PQR}$:

$\text{Area}(△\mathrm{PST})=\frac{1}{2}\text{Area}(△\mathrm{PQR})$

Substituting this ratio into our similarity relation:

$\frac{1}{2}={\left(\frac{\mathrm{PS}}{\mathrm{PQ}}\right)}^2$

Taking the square root on both sides:

$\frac{\mathrm{PS}}{\mathrm{PQ}}=\frac{1}{\sqrt{2}}$

Thus, $\mathrm{PQ}=\sqrt{2}·\mathrm{PS}$.

Step 4: Calculate the Ratio of PS to QS

We know that $\mathrm{PQ}=\mathrm{PS}+\mathrm{QS}$, so:

$\mathrm{QS}=\mathrm{PQ}-\mathrm{PS}$

$\mathrm{QS}=\sqrt{2}·\mathrm{PS}-\mathrm{PS}=\mathrm{PS}(\sqrt{2}-1)$

Now, find the ratio $\frac{\mathrm{PS}}{\mathrm{QS}}$:

$\frac{\mathrm{PS}}{\mathrm{QS}}=\frac{\mathrm{PS}}{\mathrm{PS}(\sqrt{2}-1)}=\frac{1}{\sqrt{2}-1}$

Therefore, the ratio of $\mathrm{PS}$ to $\mathrm{QS}$ is $1:\sqrt{2}-1$.