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Major Axis of Ellipse

On the Argand...

Question

On the Argand plane $z_{1}, z_{2}$ and $z_{3}$ are respectively, the vertices of an isosceles triangle ABC with AC = BC and equal angles are $θ$ . If $z_{4}$ is the incenter of the triangle.
Then prove that $(z_{2} - z_{1}) (z_{3} - z_{1}) = (1 + s e c θ) (z_{4} - z_{1})^{2}$ .

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Solution

$\frac{z_{2} - z_{1}}{| z_{2} - z_{1} |} = \frac{z_{4} - z_{1}}{| z_{4} - z_{1} |} e^{- i θ / 2}$ (clockwise) (1)
$\frac{z_{3} - z_{1}}{| z_{3} - z_{1} |} = \frac{z_{4} - z_{1}}{| z_{4} - z_{1} |} e^{- i θ / 2}$ (anticlockwise) (2)
Multiplying (1) and (2), we get
$\frac{(z_{2} - z_{1}) (z_{3} - z_{1})}{(z_{4} - z_{1})^{2}} = \frac{| (z_{2} - z_{1}) | | (z_{3} - z_{1}) |}{{| z_{4} - z_{1} |}_{4}^{2}}$
$= \frac{(A B) (A C)}{(A I)^{2}}$
$= \frac{2 (A D) (A C)}{(A I)^{2}}$
$= \frac{(A B)^{2}}{(A I)^{2}} \frac{A C}{A D}$
$= 2 c o s^{2} \frac{θ}{2} s e c θ = (1 + c o s θ) s e c θ$
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