Question

In a right angle triangle right angled at



b. bd is perpendicular to ac prove bd^2=ad.cd

Answer 1

In triangles BAD and CBD, 
angle ADB = angle CDB                [90 degrees]      (A)
angle BAD = angle DBC                     (A)

Thus, triangle BAD is similar to triangle CBD by AA criterion
=> BD/CD = AD/BD                      [cpst]
On cross-multiplication,
=> BD2 = DC.AD