In the given Fig. if AD ⊥ BC and Prove that ∆ABC is a right triangle.
Given: in which
and
To Prove:
∆ABC is a right triangle
Proof : In ∆ABD, we have
AB2 = AD2 + BD2 ...(i)
[Using Pythagoras theorem]
In ∆ACD, we have
AC2 = AD2 + CD2 ...(ii)
[Using Pythagoras theorem]
Adding (i) and (ii)
AB2 + AC2 = AD2 + BD2 + AD2 + CD2
⇒ AB2 + AC2 = 2AD2 + BD2 + CD2
⇒ AB2 + AC2 = 2(BD.DC) + BD2 + CD2
⇒ AB2 + AC2 = (BD + DC)2
⇒ AB2 + AC2 = BC2
⇒ AB2 + AC2 = BC2
Therefore, ∆ABC is a right triangle, right angled at A.