Question

From given figure, In ∆ABC, AD ⊥ BC, then prove that AB² + CD² = BD² + AC² by completing activity.

Activity: From given figure, In ∆ACD, By pythagoras theorem

AC² = AD² + `square`

∴ AD² = AC² – CD²   ...(I)

Also, In ∆ABD, by pythagoras theorem,

AB² = `square` + BD²

∴ AD² = AB² – BD²   ...(II)

∴ `square` – BD² = AC² – `square`

∴ AB² + CD² = AC² + BD²

Answer 1

From given figure, in ∆ACD, By pythagoras theorem

AC² = AD² + \[\boxed{CD^2}\]

∴ AD² = AC² – CD²   ...(I)

Also, In ∆ABD, by pythagoras theorem,

AB² = \[\boxed{AD^2}\] + BD²

∴ AD² = AB² – BD²   ...(II)

∴ \[\boxed{AB^2}\] − BD² = AC² − \[\boxed{CD^2}\]   ...[From (i) and (ii)]

∴ AB² + CD² = AC²+ BD²