Question

Please solve this.Seg AN and seg CM are the median of triangle ABC in which angleB =90॰. Prove that 4(AN square+CM square)=5AC square.
(THE DIAGRAM IS NOT GIVEN. PLEASE DRAW THE DIAGRAM YOURSELF)​ANSWER WILL BE MARKED AS BRAINLIEST ​

Answer 1

Given :

Since , CM is a median : BM = AM

and AN is a median : BN = CN

<B = 90°

Solution :

By Pythagoras Theorem ,,,

In ∆ABC ,

${AC}^{2} = {AB}^{2} + {BC}^{2}$

equation ( 1 )

In ∆BMC ,

${CM}^{2} = {BM}^{2} + {BC}^{2}$

${CM}^{2} = { \frac{1}{2}AB }^{2} + {BC}^{2}$

${CM}^{2} = \frac{ {AB}^{2} }{4} + {BC}^{2}$

${CM}^{2} = \frac{ {AB}^{2} +4 {BC}^{2} }{4}$

$4 {CM}^{2} = {AB}^{2} + 4 {BV}^{2}$

equation ( 2 )

In ∆ABN ,

${AN}^{2} = {AB}^{2} + {BN}^{2}$

${AN}^{2} = {AB}^{2} + { \frac{1}{2}BC }^{2}$

${AN}^{2} = {AB}^{2} + \frac{ {BC}^{2} }{4}$

${AN}^{2} = \frac{4 {AB}^{2} + {BC}^{2} }{4}$

$4 {AN}^{2} = 4 {AB}^{2} + {BC}^{2}$

equation ( 3 )

Add equation ( 3 ) and ( 2 ) ,,,,

${4AN}^{2} + 4 {CM}^{2} = 4 {AB}^{2} + {BC}^{2} + {AB}^{2} + 4 {BC}^{2}$

$4( {AN}^{2} + {CM}^{2} ) = 5 {AB}^{2} + 5 {BC}^{2}$

$4( {AN}^{2} + {CM}^{2} ) = 5( {AB}^{2} + {BC}^{2} )$

$4( {AN}^{2} + {CM}^{2} ) = 5 {AC}^{2}$

Hope it's helpful ......❤