If △ABC is right-angled atB. then prove that AB2+BC2=AC2
We know that if a perpendicular is drawn from the vertex of a right angle of a right angled triangle to the hypotenuse, then triangle on both sides of the perpendicular are similar to the whole triangle and to each other.
We have
△ADB∼△ABC
∴ADAB=ABAC (In similar triangles corresponding sides are equal)
AB2=AD×AC.........(1)
Also △BDC∼△ABC
∴CDBC=BCAC (In similar triangles corresponding sides are equal)
BC2=CD×AC.........(2)
Adding (1) and (2) we get
AB2+BC2=AD×AC+CD×AC
AB2+BC2=AC(AD+A=CD)
From the figure AD+CD=AC
AB2+BC2=AC.AC
Therefore, AB2+BC2=AC2