Question

Using the conditions from 32(b), prove that \(PR = 2AP\).

Hint:

The ratio of corresponding sides of similar triangles is constant. Always relate the base segments (\(AD\) and \(BR\)) first.

Answer 1

Step 1: Understanding the Concept:
This follows directly from the similarity of triangles established in part (ii).
Step 3: Detailed Explanation:
1. As shown in part (ii), \(\triangle APD \sim \triangle RPB\).
2. The ratio of their sides is:
\[ \frac{AP}{PR} = \frac{AD}{BR} \]
3. Since \(ABCD\) is a parallelogram, \(AD = BC\).
4. From the congruence \(\triangle ADQ \cong \triangle RCQ\), we have \(AD = CR\).
5. Point \(R\) is on the extension of \(BC\), so \(BR = BC + CR = AD + AD = 2AD\).
6. Substitute this back into the similarity ratio:
\[ \frac{AP}{PR} = \frac{AD}{2AD} = \frac{1}{2} \]
7. Cross-multiplying gives:
\[ PR = 2AP \]
Step 4: Final Answer:
The similarity ratio \(\frac{AP}{PR} = \frac{1}{2}\) leads directly to \(PR = 2AP\).