Question

in an obtuse triangle ABC angle C is an obtuse angle AD perpendicular to BC meats BC at D on extending forward prove that AB sq. = ACsq.+BCsq.+2BC ×CD

Answer 1

now use Pythagoras theorem,

$AB^{2} = AD^{2} + BD^{2}$ .....(1)

we know that BD = CB+CD so,

$AB^{2}= AD^{2}+ (CB+CD) ^{2}$

$= AD^{2} + CB^{2} + CD^{2} + 2 (BC(CD))$ ......(2)

now we know that $AD^{2}= AC^{2} - CD^{2}$

now  substitute it in (2)

$AB^{2} = AC^{2} - CD^{2} + BC^{2} +CD ^{2} + 2 (BC(CD))$

hence proved