Question

in a isosceles triangle ABC,AB=ACand D is a point on BC produced. prove that ADsquare =ACsquare+BD.CD

Answer 1

In order to draw the result ,first draw AE  ⊥  BC

Perpendicular drawn from the vertex to the opp. base of an isosceles Δ bisects the base 
∴BE = EC 

Applying pythagoras theorem in right Δ AED
AD² = AE² + ED².....................(i)

Again applying pythagoras theorem in right Δ AEC
AC² = AE² + EC²...............(ii)

From (i) and (ii) ,we get
AD² = (AC² - EC²) + ED²               [ AC² = AE² + EC² ⇒ AE² = AC² - EC²]

AD² = AC² + (- EC² + ED² )
AD² = AC² + ( ED² - EC² )
AD² = AC² + ( ED + EC )( ED - EC)
AD² = AC² + ( ED + BE )( ED - EC)                {∵BE = EC }

⇒AD² = AC² + BD.CD