Question

In the given figure, P and Q are mid points of the sides BC and CA respectively of a triangle ABC, right angled at C. Prove that: 1- 4AP^2 = 4AC^2 + BC^2 2- 4BQ^2=4BC^2+AC^2 3- 4(AP^2+BQ^2) = 5AB^2 PLSSS FASTTT I WILL MARK BRAINLIEST

Answer 1

Answer:

$</p><p></p><p>\begin{gathered}{aq}^{2} = {ac}^{2} + {cq}^{2} \\ = > {aq}^{2} = {ac}^{2} + { (\frac{bc}{2} )}^{2} \\ = > {aq}^{2} = {ac}^{2} + \frac{bc}{4} \\ = > {aq}^{2} = \frac{4 {ac}^{2} + {bc}^{2} }{4}\end{gathered}aq2=ac2+cq2=>aq2=ac2+(2bc)2=>aq2=ac2+4bc=>aq2=44ac2+bc2</p><p></p><p>= > {4aq}^{2} = {4ac}^{2} + {bc}^{2}=>4aq2=4ac2+bc2</p><p></p><p>$

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