Question

S is any point on the side QR of a \(\Delta PQR\) such that \(\angle PSR = \angle QPR\). Prove that \(\frac{QR}{RP} = \frac{RP}{RS}\).

• A. Proof Completed
• B. Triangles Not Similar
• C. Insufficient Data
• D. Cannot be Determined

Hint:

To avoid mistakes in writing the ratios of similar triangles, always list the triangles with their corresponding vertices in order:
\[ \Delta QPR \sim \Delta PSR \]
The ratios can then be written directly from the order of letters:
\[ \frac{QP}{PS} = \frac{PR}{SR} = \frac{QR}{PR} \]
This eliminates any visual confusion from the diagram.

Answer 1

The Correct Option is A

Step 1: Understand what is being asked.
In $\Delta PQR$, $S$ lies on $QR$ with $\angle PSR=\angle QPR$, and we must prove $\frac{QR}{RP}=\frac{RP}{RS}$. Instead of writing the similarity ratio directly in this final form, let us first prove the equivalent geometric mean relation $RP^2=RQ\times RS$, and then rearrange it.

Step 2: Prove the two triangles $\Delta RPS$ and $\Delta RQP$ are similar.
In these two triangles:
$\angle PRS$ and $\angle QRP$ are the same angle, since $S$ lies on segment $QR$, so they share the vertex $R$ and the same pair of rays.
$\angle PSR = \angle QPR$ is given directly in the problem.
By AA similarity (two equal angles):
\[ \Delta RPS \sim \Delta RQP \]
Step 3: Write the geometric mean relation from this similarity.
Matching corresponding sides ($R\to R$, $P\to Q$, $S\to P$ in the correspondence $RPS\sim RQP$), the sides opposite equal angles give:
\[ \frac{RP}{RQ} = \frac{RS}{RP} \]
Cross-multiplying gives the geometric mean form:
\[ RP^2 = RQ\times RS \]
Step 4: Rearrange this into the required ratio form.
Starting from $RP^2=RQ\times RS$, divide both sides by $RP\times RS$:
\[ \frac{RP^2}{RP\times RS} = \frac{RQ\times RS}{RP\times RS} \]
\[ \frac{RP}{RS} = \frac{RQ}{RP} \]
which is exactly $\frac{QR}{RP}=\frac{RP}{RS}$ (writing $RQ$ as $QR$, the same segment).

Final Answer:
The relation $\frac{QR}{RP}=\frac{RP}{RS}$ is proved using the similarity of $\Delta RPS$ and $\Delta RQP$.
\[ \boxed{\frac{QR}{RP}=\frac{RP}{RS}} \]