Question

in the figure triangle ABC is a right angled triangle at c and d is the midpoint BC. prove that ab square= 4ad square -3ac square

Answer 1

$here \: is \: your \: answer \: mate$

$in \: triangle \: abc \\ {ab}^{2} = {ac}^{2} + {cb}^{2}$

${ab}^{2} = {ad}^{2} - {cd }^{2} + {bc}^{2} \\ {ab}^{2} = {ad}^{2} - { \frac{bc}{4} }^{2} + {bc}^{2} \\ {ab}^{2} = \frac{ {4ad \:}^{2} - {bc}^{2} + {4bc}^{2} }{4}$

${4ab}^{2} = {3ac}^{2} + {4ad}^{2}$

$here \: is \: your \: answer$