Let’s solve the problem step by step based on the image provided.

Problem Breakdown:

The figure shows graphs of $y = b^x$ and $y = c^x$ on a coordinate system, where $b$ and $c$ are positive constants. A horizontal line $L$ intersects both graphs, producing points $A, B,$ and $C$. We are tasked to determine which statements about $b$ and $c$ are true:

Statements:

1. $b > c$
2. $b + c > 2$
3. $\frac{AB}{AC} = \log_c b$

Solution:

Step 1: Analyze Statement I

• $y = b^x$ grows faster than $y = c^x$ (i.e., it is above $y = c^x$ for $x > 0$), indicating that $b > c$.
• Conclusion: Statement $I$ is true.

Step 2: Analyze Statement II

• $b$ and $c$ are positive constants, and we know $b > c$. Since $b > 1$ (for an exponential graph that grows), and $c > 0$, it follows that $b + c > 2$.
• Conclusion: Statement $II$ is true.

Step 3: Analyze Statement III

• $AB$ and $AC$ are distances on the $y$-axis, given as: $AB = k - b^0 = k - 1, \quad AC = k - c^0 = k - 1$ where $k$ is the value of the horizontal line $L$.

• Therefore: $\frac{AB}{AC} = \frac{k - 1}{k - 1} = 1$

• The given equation $\frac{AB}{AC} = \log_c b$ implies: $\log_c b = 1$ This is incorrect because $\log_c b \neq 1$ unless $b = c$, which contradicts $b > c$.

• Conclusion: Statement $III$ is false.

Final Answer:

The correct option is A. I and II only.

Would you like a further breakdown of any step? Here are some related questions to expand on this concept:

1. How can we verify $b + c > 2$ with different values of $b$ and $c$?
2. Why does $y = b^x$ grow faster than $y = c^x$ when $b > c$?
3. How does the logarithmic equation $\log_c b$ relate to this problem geometrically?
4. What happens if $b = c$? Would the statements still hold true?
5. How can we generalize this problem for any exponential functions $y = a^x$ and $y = d^x$?

Tip: For exponential functions $y = a^x$, the base $a$ determines the growth rate. A higher base means faster growth for $x > 0$.