Question 3

We will use the Basic Proportionality Theorem (Thales's Theorem) in two different triangles.

Step 1 — Using BPT in triangle ABC

We are given that line segment $\text{LM}$ is parallel to line segment $\text{CB}$. Consider triangle $\text{ABC}$. Since $\text{LM} || \text{CB}$, we can apply the Basic Proportionality Theorem. The theorem states that a line parallel to one side of a triangle divides the other two sides proportionally. So, we have the ratio:

$\frac{\text{AM}}{\text{AB}} = \frac{\text{AL}}{\text{AC}} \quad \text{(Equation 1)}$

Step 2 — Using BPT in triangle ADC

We are given that line segment $\text{LN}$ is parallel to line segment $\text{CD}$. Consider triangle $\text{ADC}$. Since $\text{LN} || \text{CD}$, we can apply the Basic Proportionality Theorem again. The line $\text{LN}$ divides sides $\text{AD}$ and $\text{AC}$ proportionally. So, we have the ratio:

$\frac{\text{AN}}{\text{AD}} = \frac{\text{AL}}{\text{AC}} \quad \text{(Equation 2)}$

Step 3 — Comparing the results

Now, let's look at Equation 1 and Equation 2. Both equations have $\frac{\text{AL}}{\text{AC}}$ on their right-hand side. This means their left-hand sides must be equal to each other. From Equation 1, we have $\frac{\text{AM}}{\text{AB}} = \frac{\text{AL}}{\text{AC}}$. From Equation 2, we have $\frac{\text{AN}}{\text{AD}} = \frac{\text{AL}}{\text{AC}}$. Therefore, we can conclude:

$\frac{\text{AM}}{\text{AB}} = \frac{\text{AN}}{\text{AD}}$

Answer

$\boxed{\frac{\text{AM}}{\text{AB}} = \frac{\text{AN}}{\text{AD}}}$