Question

Plz solve 25th question

Answer 1

in triangle ADC

hyp2 +base2 + hight2

${ac}^{2} = {dc}^{2} + {da}^{2}$

therefore db=dc

so it can be written

${ac}^{2} = {da}^{2} + {{bc +db}^{2}$

${ac}^{2} = {da}^{2} + ( {bc}^{2} + {db}^{2} + 2.bc.bd)$

now in triangle ABD

${ab}^{2} = {db}^{2} + {da}^{2}$

there fore in place of

${da}^{2} and \: {db}^{2} \: we \: can \: write \: \: {ab}^{2}$

so

${ac}^{2} = {ab}^{2} + 2.bc.bd + {bc}^{2}$

hence proved

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Answer 2

here is your answer
hope this will help you
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