Question

AD is the altitude of ΔABC such that B-D-C. If AD² = BD·DC, prove that ∠BAC is right angle.
[Hint : AD⊥BC. So, B-D-C is given. So, ΔADB and ΔADC are right angled triangles to which Pythagoras' theorem can be applied.

Answer 1

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 AD2 = BD.DC …… (3)
From equation (1) and (3), we get,
AB2 = BD2 + BD.DC
⇒ AB2 = BD(BD + DC) = BD. BC.
$\implies BD = \frac{AB^2}{BC}$
now from equations (2) and (3),
AC² = BD.DC + DC²
⇒AC² = DC(BD + DC)
⇒ AC² = DC.BC
$\implies DC=\frac{AC^2}{BC}$
Also, BC = BD + DC,
Substituting the value of BD and DC in above equation, we get,
$\implies BC=\frac{AB^2}{BC}+\frac{AC^2}{BC}$
⇒ BC2 = AB2 + AC2
So, by the converse of Pythagoras Theorem, we can say that ∠BAC is a right angle.