To prove the equation \( b = c \sin A + a \sin C \) geometrically, let's break it down step by step using the context of a triangle.

### Step 1: Consider a triangle
We are working within a triangle, say \( \triangle ABC \), where:
- \( A \), \( B \), and \( C \) are the angles.
- \( a \), \( b \), and \( c \) are the sides opposite to the angles \( A \), \( B \), and \( C \) respectively.

The goal is to prove that:
\[
b = c \sin A + a \sin C
\]

### Step 2: Set up the triangle and altitudes
Consider an arbitrary triangle \( \triangle ABC \) with vertices \( A \), \( B \), and \( C \), and the sides labeled as mentioned before. Let us draw perpendiculars (altitudes) from vertex \( B \) onto sides \( AC \). 

Let's denote the height from vertex \( B \) to line \( AC \) as \( h \). This gives two right triangles:
1. \( \triangle BCA \) 
2. \( \triangle BAC \)

### Step 3: Apply the sine rule
In any triangle, the sine of an angle is the ratio of the opposite side to the hypotenuse. Using this for angles \( A \) and \( C \) in their respective right triangles, we get the following relationships:

1. From \( \triangle BCA \):
\[
h = c \sin A
\]
2. From \( \triangle BAC \):
\[
h = a \sin C
\]

### Step 4: Summing the expressions
Since both expressions represent the height \( h \), we can set them equal:
\[
h = c \sin A + a \sin C
\]

Finally, notice that this is the same

Discover how to geometrically prove the equation b = c sin A + a sin C using triangle altitudes and the sine rule. This step-by-step guide breaks down the problem for students in Grades 10-12, covering key trigonometric concepts.

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