Question

In \( \triangle ABC \), \( \angle C = 90^\circ \) and \( P, Q \) are midpoints of \( CA \) and \( CB \) respectively. Prove that \( 4AQ^2 = 4AC^2 + BC^2 \).

Hint:

Midpoint Theorem in Right Triangle: If P, Q are midpoints of right triangle sides, use \( 4AQ^2 = 4AC^2 + BC^2 \).

Answer 1

Since \( P \) and \( Q \) are midpoints,
\[ AQ = \frac{1}{2} AB, \quad PQ = \frac{1}{2} BC \] Applying the midpoint theorem:
\[ 4AQ^2 = AB^2 + 4PQ^2 \] Since \( AB^2 = AC^2 + BC^2 \) (Pythagoras theorem),
\[ 4AQ^2 = 4AC^2 + BC^2 \] Thus, the given equation is proved.
Correct Answer: \( 4AQ^2 = 4AC^2 + BC^2 \) is proved.