In triangle ABC is right triangle with angle B=90 degree . P and Q are midpoints of AB and BC. then prove that: (i) 4 PCÂ²=4 BCÂ² + ABÂ²(ii) 4 AQÂ² = 4 ABÂ² + BCÂ²(iii) 4 PCÂ² + 4 AQÂ² = 5 ACÂ²

$Given\;that\;AB=2PB=2AP\;and\;BC=2BQ=2QC\\*(1)\;4PC^2=4BC^2+AB^2\\*In\;\triangle PBC\\*LHS=4PC^2=4(PB^2+BC^2) \\*\Rightarrow 4(\frac{AB^2}{4}+BC^2)\\*\Rightarrow 4BC^2+AB^2\\* Hence,\;proved\;4PC^2=4BC^2+AB^2\\* (2)\;4AQ^2=4AB^2+BC^2\\*In\;triangleABQ\\*LHS=4AQ^2=4(AB^2+BQ^2)\\*\Rightarrow 4(AB^2+\frac{BC^2}{4})\\* \Rightarrow 4AB^2+BC^2\\* Hence,\;proved\;4AQ^2=4AB^2+BC^2\\* (3)\;4PC^2+4AQ^2=5AC^2\\*LHS=4PC^2+4AQ^2\\*\Rightarrow 4(BC^2+PB^2)+4(AB^2+BQ^2)\\* \Rightarrow 4(BC^2+\frac{AB^2}{4})+ 4(AB^2+\frac{BC^2}{4})\\* \Rightarrow 4BC^2+AB^2+BC^2+4AB^2\\* \Rightarrow 5(AB^2+BC^2)\\*\therefore In\;\triangle ABC\\* AC^2=AB^2+BC^2\\* \Rightarrow 5(AC^2)\\* Hence,\;proved\;4PC^2+4AQ^2=5AC^2$

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