Q4 of 53 Page 158

In ΔABC, , B—D—C. If AC2 = CD ⋅ BC, prove that ∠BAC is a right angle.


Since AD is the altitude to BC,


⇒ ∠BDA = ∠CDA = 90°


So, Δ ABD and Δ ACD are right angle triangles.


Using Pythagoras theorem,


AB2 = BD2 + AD2 ...... (1)


And, AC2 = AD2 + DC2 ...... (2)


It is given that AC2 = CD.BC



From equation (1) and (2), we get,


AB2 – BD2 = AC2 – CD2


⇒ AB2 – (BC – CD)2 = AC2 – CD2


⇒ AB2 – BC2 – CD2 + 2.BC.CD = AC2 – CD2


⇒ AB2 – BC2 + 2.BC.CD = AC2


Substituting the value of CD above, we get,


⇒ AB2 – BC2 + 2AC2 = AC2


⇒ AB2 + AC2 = BC2


So, by the converse of Pythagoras Theorem, we can say that ∠BAC is a right angle.


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