Triangle $ABC$ is acute. Point $D$ lies on $\overline {AC}$ so that $\overline {BD}\perp \overline {AC}$, and point $E$ lies on $\overline {AB}$ such that $\overline {CE}\perp\overline {AB}$. The intersection of segments $\overline {CE}$ and $\overline {BD}$ is $P$. Find the value of $AE$ if $CP=10$, $PE=20$, and $EB=30$.

To solve this problem, we first understand the given geometry and employ coordinate geometry and length relationships.

### Step 1: Understanding the Setup
We are given the following:
- $ \triangle ABC $ is acute.
- Point $ D $ is on $ \overline{AC} $ such that $ \overline{BD} \perp \overline{AC} $.
- Point $ E $ is on $ \overline{AB} $ such that $ \overline{CE} \perp \overline{AB} $.
- Segments $ \overline{CE} $ and $ \overline{BD} $ intersect at point $ P $.
- Given lengths: $ CP = 10 $, $ PE = 20 $, and $ EB = 30 $.

We are required to find the length of $ AE $.

### Step 2: Use Length Ratios in Similar Triangles
We start by recognizing that since $ \overline{BD} $ and $ \overline{CE} $ are altitudes, $ P $ is the orthocenter of $ \triangle ABC $. 

Now, let's analyze the triangles formed by the intersection point $ P $.

### Step 3: Consider Similar Triangles
1. Triangles $ \triangle CEP $ and $ \triangle BEP $:
   - $ \triangle CEP \sim \triangle BEP $ since they share angle $ \angle EBP $ and have right angles at $ E $.

2. Set up the ratio of the sides using the similarity:
   \[
   \frac{CP}{PE} = \frac{CE}{EB}.
   \]
   Plugging in the known values:
   \[
   \frac{10}{20} = \frac{CE}{30}.
   \]
   Simplify:
   \[
   \frac{1}{2} = \frac{CE}{30}.
   \]
   Therefore, 
   \[
   CE = 15.
   \]

3. Use the length $ CE $ to find $ AE $. Since $ \triangle AEC $ shares point $ E $ on $ AB $ and point $ C $ on $ AC $, we can use the section formula:
   
   Since $ AE + EB = AB $, and given $ EB = 30 $, we need to find $ AE $. 

4. Use a relationship between $ PE $ and $ AE $:
   Since $ P $ divides $ CE $ in a $ 1:2 $ ratio (because $ CP = 10 $ and $ PE = 20 $), it also implies that $ AE = \frac{1}{2} \times 30 = 10 $. 

Thus, the length of $ AE $ is:

\[
\boxed{10}
\] 

### Conclusion:
The value of $ AE $ is $ 10 $.

Would you like to see more details about this or have any further questions?

### Related Questions:
1. What are the properties of the orthocenter in an acute triangle?
2. How would the solution change if the triangle were obtuse?
3. How can coordinate geometry be used to solve this problem differently?
4. What is the significance of similar triangles in this setup?
5. Can you find the area of $ \triangle ABC $ given these lengths?

**Tip:** Always check for relationships like similar triangles when you have perpendicular lines intersecting.

Triangle $ABC$ is acute. Point $D$ lies on $\overline{AC}$ so that $\overline{BD} \perp \overline{AC}$, and point $E$ lies on $\overline{AB}$ such that $\overline{CE} \perp \overline{AB}$. The intersection of segments $\overline{CE}$ and $\overline{BD}$ is $P$. Find the value of $AE$ if $CP=10$, $PE=20$, and $EB=30$.

Solve for AE in triangle $ABC$ where $BD$ and $CE$ are altitudes, and segments $\overline{CE}$ and $\overline{BD}$ intersect at $P$. Given $CP=10$, $PE=20$, and $EB=30$, learn how to apply theorems on similar triangles and orthocenters to find $AE$. Suitable for Grades 9-12.

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