Question

If A (z₁), B(z₂), C(z₃) are the vertices of a triangle ABC inscribed in the circle |z| = 2. Internal angle bisector of the angle A, meet the circum-circle again at D (z₄) then –

• A. z₄² = z₂z₃
• B. z₄ = $\frac{z_{2}z_{3}}{z_{1}}$
• C. z₄ = $\frac{z_{1}z_{2}}{z_{3}}$
• D. z₄ = $\frac{z_{1}z_{3}}{z_{2}}$

Answer 1

Answer: (A)

z₄² = z₂z₃

Answer 2

Sol.

Now Ð BOD = 2 Ð BAD = A

and Ð COD = 2 ÐCAD = A

from the rotation about the point O, we get

$\frac{z_{4}}{z_{2}}$= e^iA, $\frac{z_{3}}{z_{4}}$= e^iA Þ z₄² = z₂ z₃