Question

In ∆ abc, If AB=AC and D is on a point BC.Prove that AB^2-AD^2= BD×CD.​

Answer 1

Answer:

QED

Step-by-step explanation:

lets draw an perpendicular AE from point A to at BC

as ab = ac so triangle is isosceles

so BE = CE = BC/2    and AE is perpendicular to BC

in triangle  ABE

AB^2 = AE^2 + BE^2    

AD^2 = AE^2 + DE^2

AB^2 - AD^2 = BE^2 - DE^2

AB^2 - AD^2 = (BE+DE)(BE-DE)

assuming D is on left to point E

as BE = CE

AB^2 - AD^2 =(CE + DE)(BE-DE)

AB^2 - AD^2 = CD * BD

assuming D is on right to point E

AB^2 - AD^2 =(BE + DE)(CE-DE)

AB^2 - AD^2 =BD*CD

if D & E are same points then DE = 0

AB^2 - AD^2 = (CE)(BE)

CE = BE = CD = BD

AB^2 - AD^2 = (CD)(BD)

QED

Step-by-step explanation: