Question

Annyeonghaseyo guys!!
Plz solve this question quick.
N no spamming for God's sake!

Answer 1

In right-triangle ABD,
by pythagoras theorem,

${ab}^{2} = {ad}^{2} + {bd}^{2} \: \: \: \: \: \: eq.1$
Now, in right-triangle ACD,
by pythagoras theorem,
${ad}^{2} = {ac}^{2} - {cd}^{2} \: \: \: \ \: \: \: eq.2$

Now we can see that,
BD = BC + CD

On squaring both sides,
${ bd}^{2} = {(bd + cd)}^{2} \\ {bd}^{2} = {bd}^{2} + {cd}^{2} + 2.bc.cd \: \: \: \: \: \: \: \: eq.3$

Now put values of
${ad}^{2} \: \: \: \: and \: \: \: \: {bd}^{2}$
from eq. 2 and eq. 3 in eq. 1.
${ab}^{2} = {ad}^{2} + {cd}^{2} \\ {ab}^{2} = {ac}^{2} - {cd}^{2} + {bc}^{2} + {cd}^{2} + 2.bc.cd \\ {ab}^{2} = {bc}^{2} + {ac}^{2} + {cd}^{2} - {cd}^{2} + 2.bc.cd \\ {ab}^{2} = {bc}^{2} + {ac}^{2} + 2.bc.cd$
Hence proved.