https://file.aiquickdraw.com/mathbot/file/1727172682461loami6zm.jpg

Let's go through the two geometry problems presented in the image.

### Problem 26 (Part A):
**In triangle \( \triangle ABC \), D, E, and F are midpoints of BC, CA, and AB respectively. Prove that:**
1. \( \triangle FBD \sim \triangle DEF \)
2. \( \triangle DEF \sim \triangle ABC \)

**Solution:**

1. **Proving \( \triangle FBD \sim \triangle DEF \):**
   - Given that \( D \), \( E \), and \( F \) are midpoints, segment \( DE \parallel AB \), segment \( DF \parallel AC \), and segment \( EF \parallel BC \) by the **Midpoint Theorem**.
   - Since \( EF \parallel BC \), and both triangles \( FBD \) and \( DEF \) share angle \( \angle FBD = \angle DEF \), this gives us **corresponding angles**.
   - \( BD \parallel EF \) implies that \( \triangle FBD \) and \( \triangle DEF \) have two pairs of corresponding angles equal, which by the **AA similarity criterion** implies \( \triangle FBD \sim \triangle DEF \).

2. **Proving \( \triangle DEF \sim \triangle ABC \):**
   - Since \( D, E, F \) are midpoints, \( DE \parallel AB \), \( DF \parallel AC \), and \( EF \parallel BC \), the triangles \( \triangle DEF \) and \( \triangle ABC \) are similar by **AA similarity criterion** (corresponding angles are equal and the sides are proportional in a 1:2 ratio).

### Problem 26 (Part B):
**In triangle \( \triangle ABC \), P and Q are points on AB and AC, respectively, such that \( PQ \parallel BC \).**

This problem involves the concept of similarity in triangles. Since \( PQ \parallel BC \), the triangles \( \triangle APQ \) and \( \triangle ABC \) are similar by the **Basic Proportionality Theorem (Thales Theorem)**. This gives proportionality between the sides, and \( \triangle APQ \sim \triangle ABC \).

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### Problem:
**Prove that the median AD drawn from A on BC bisects PQ.**

**Solution:**
- We are given that \( PQ \parallel BC \).
- Since \( PQ \parallel BC \), and \( AD \) is the median of \( \triangle ABC \), it implies that triangles \( \triangle APD \) and \( \triangle AQC \) are congruent by **SAS congruence** (they share a common angle \( \angle A \), and sides \( AP = AQ \) as well as \( PD = DQ \) since \( AD \) is the median).
- Since \( AD \) bisects \( BC \) and \( PQ \parallel BC \), the median \( AD \) must also bisect \( PQ \). Therefore, \( AR = RP \).

Would you like me to go over any steps in more detail?

### Relative Questions:
1. How does the Midpoint Theorem help in proving triangle similarity?
2. What are the key criteria for proving triangles are similar?
3. How does the Basic Proportionality Theorem (Thales' Theorem) apply in triangle geometry?
4. What is the role of parallel lines in proving triangle similarity?
5. How does median bisecting a line segment relate to symmetry in a triangle?

### Tip:
In geometry problems involving triangle similarity or proportionality, always look for shared angles and parallel lines to apply the AA similarity criterion or the Basic Proportionality Theorem.

In triangle ABC, D, E, and F are midpoints of BC, CA, and AB respectively. Prove that triangles FBD ~ DEF and DEF ~ ABC. OR In triangle ABC, P and Q are points on AB and AC respectively such that PQ is parallel to BC. Prove that the median AD drawn from A on BC bisects PQ.

This geometry problem focuses on proving triangle similarity using the AA similarity criterion and midpoint theorem. It also involves showing how a median bisects a line parallel to the base of a triangle. Suitable for students in Grades 9-10.

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