Question

In triangle ABC,A=60° and ad is median,then prove 4ad²=c²+b²+bc

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Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Answer:-

We have,

$\sf \angle A=60\degree$

$\sf \implies \cos(\angle A)= \frac{1}{2}$

$\sf \implies \frac{ { {b}^{2} } + {c}^{2} - {a}^{2} }{2bc} = \frac{1}{2}$

$\sf \implies { {b}^{2} } + {c}^{2} - {a}^{2} = bc$

$\sf \implies {a}^{2} = { {b}^{2} } + {c}^{2} - bc \: ...(i)$

Applying law of cosine in $\sf\Delta ABD, we have, </p><p>[tex] \sf \implies {AD}^{2} = {c}^{2} + \frac{ {a}^{2} }{4} - 2 \times c \times \frac{a}{2} + \cos(B)$

$\sf \implies 4 {AD}^{2} = 4 {c}^{2} + {a}^{2} - 4ac \cos(B)$

$\sf \implies 4 {AD}^{2} = 4 {c}^{2} + {a}^{2} - 4ac \times \big( \frac{ {a}^{2} + {c}^{2} - {b}^{2} }{2ac} \big)$

$\sf \implies 4 {AD}^{2} = 4 {c}^{2} + {a}^{2} - 2( {a}^{2} - {c}^{2} - {b}^{2} )$

On simplifying, we get,

$\sf \implies 4 {AD}^{2} =2 {c}^{2} - {a}^{2} + 2 {b}^{2}$

From (i), we get,

$\sf \implies 4 {AD}^{2} =2 {c}^{2} - ( {b}^{2} + {c}^{2} - bc)+ 2 {b}^{2}$

$\sf \implies 4 {AD}^{2} = {b}^{2} + {c}^{2} + bc$

Hence Proved.