Question:In the figure AB = BC and AD is perpendicular to CD. Prove that AC2 = 2.BC. DC

solution :In ∆ ADC

AC^{2} = AD^{2} + DC^{2}

AC^{2} = AB^{2} – BD^{2} + DC^{2}(AS ABD is a right angled triangle)

or AC^{2} = AB^{2} + DC^{2} – BD^{2}

AC^{2} = AB^{2} + DC^{2} – ( DC-BC )^{2}

or AC^{2} = AB^{2} + DC^{2} – (DC^{2} + BC ^{2} -2 DC. BC)

or AC^{2} = AB^{2} + ~~DC~~ – ^{2} ~~ DC~~ – BC ^{2}^{2} + 2 DC. BC

or AC^{2} =~~ AB~~ –^{2} ~~ AB ~~ + 2 DC. BC (As AB = BC ) ^{2}

.’. AC^{2} = 2 DC. BC Proved