In triangle △ABC, ∠A is a right angle and BP and CQ are medians. Prove that: \[ 5BC^2 = 4(BP^2 + CQ^2) \]

Q.In triangle △ABC, ∠A is a right angle and BP and CQ are medians. Prove that: \[ 5BC^2 = 4(BP^2 + CQ^2) \]

Given: In triangle △ABC, ∠A = 90° and BP and CQ are medians. To Prove: 5BC² = 4(BP² + CQ²) Proof: From the right-angled triangle △ABC, BC² = AB² + AC² = (2AP)² + (2AQ)² = 4(AP² + AQ²) From triangle △ABP (right-angled), BP² = AB² + AP² = (2AQ)² + AP² = 4AQ² + AP² From triangle △ACQ (right-angled), CQ² = AC² + AQ² = (2AP)² + AQ² = 4AP² + AQ² Therefore, BP² + CQ² = 4AQ² + AP² + 4AP² + AQ² = 5AP² + 5AQ² = 5(AP² + AQ²) Now, 4(BP² + CQ²) = 4 × 5(AP² + AQ²) = 5 × 4(AP² + AQ²) = 5BC² Hence, 5BC² = 4(BP² + CQ²) — Proved.