In a △ ABC, AD is a median and $AE \perp BC$. Prove that: $AC^{2}=AD^{2}+BC\cdot DE+\left(\frac{BC}{2}\right)^{2}$

Step 1: Apply Pythagorean Theorem in $\triangle AEC$

Since $AE \perp BC$, $\triangle AEC$ is a right-angled triangle with the right angle at $E$. We can apply the Pythagorean theorem, which states that the square of the hypotenuse ($AC$) is equal to the sum of the squares of the other two sides ($AE$ and $EC$).

Step 2: Express $EC$ in terms of $DC$ and $DE$

From the diagram, we can see that the segment $EC$ can be expressed as the difference between $DC$ and $DE$. Since $AD$ is a median, $D$ is the midpoint of $BC$, so $DC = \frac{BC}{2}$.

Step 3: Substitute $EC$ and expand the equation

Substitute the expression for $EC$ from the previous step into the Pythagorean equation. Then, expand the term $(DC - DE)^2$ using the algebraic identity $(a-b)^2 = a^2 - 2ab + b^2$.

Step 4: Apply Pythagorean Theorem in $\triangle AED$

Similarly, $\triangle AED$ is also a right-angled triangle with the right angle at $E$. Applying the Pythagorean theorem to this triangle, we get $AD^2 = AE^2 + DE^2$.

Step 5: Substitute $AD^2$ and $DC = \frac{BC}{2}$

Now, substitute $AD^2$ for $(AE^2 + DE^2)$ in the expanded equation. Also, replace $DC$ with $\frac{BC}{2}$ since $D$ is the midpoint of $BC$. This simplifies the equation to the desired form.

Step 6: Rearrange to match the required form

Rearrange the terms to match the exact form required in the problem statement. This completes the proof.

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