Question

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

Hint:

Express all segments in terms of one major segment (like \(AQ\)) to compare them easily.

Answer 1

Step 1: Understanding the Concept:
We will use the similarity of triangles \(\triangle APD\) and \(\triangle RPB\) and the midpoint property.
Step 3: Detailed Explanation:
1. In \(\triangle APD\) and \(\triangle RPB\):
- \(\angle PAD = \angle PRB\) (Alternate interior angles, \(AD \parallel BR\)).
- \(\angle ADP = \angle RBP\) (Alternate interior angles, \(AD \parallel BR\)).
- Thus, \(\triangle APD \sim \triangle RPB\) by AA similarity.
2. Therefore, the ratio of corresponding sides is equal:
\[ \frac{AP}{PR} = \frac{AD}{BR} \]
3. We know \(BR = BC + CR\).
In parallelogram \(ABCD\), \(AD = BC\).
From part (i), \(\triangle ADQ \cong \triangle RCQ \implies AD = CR\).
So, \(BR = AD + AD = 2AD\).
4. Substitute \(BR = 2AD\) into the ratio:
\[ \frac{AP}{PR} = \frac{AD}{2AD} = \frac{1}{2} \implies PR = 2AP \]
5. From part (i), \(Q\) is the midpoint of \(AR\), so \(AR = 2AQ\).
Also, \(AR = AP + PR = AP + 2AP = 3AP\).
So, \(3AP = 2AQ \implies AP = \frac{2}{3}AQ\).
6. Now, find \(PQ\):
\[ PQ = AQ - AP = AQ - \frac{2}{3}AQ = \frac{1}{3}AQ \]
7. Compare \(AP\) and \(PQ\):
\[ AP = \frac{2}{3}AQ = 2 \times \left( \frac{1}{3}AQ \right) = 2PQ \]
Step 4: Final Answer:
Since \(AP = \frac{2}{3}AQ\) and \(PQ = \frac{1}{3}AQ\), it follows that \(AP = 2PQ\).