If the vertices A, B and C of an isosceles tr | vedclass

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(C) Since $	riangle ABC$ is an isosceles right-angled triangle with $\angle C = 90^{\circ}$,we have $AC = BC$. Using the property of rotation,the vec

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$(z_1-z_2)^2=2(z_1-z_3)(z_3-z_2)$

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(C) Since $	riangle ABC$ is an isosceles right-angled triangle with $\angle C = 90^{\circ}$,we have $AC = BC$. Using the property of rotation,the vector $\vec{CA}$ is obtained by rotating $\vec{CB}$ by $90^{\circ}$ ($i.e., \frac{\pi}{2}$ radians) counter-clockwise. Thus,$z_1 - z_3 = i(z_2 - z_3)$. Also,since it is isosceles,$|z_1 - z_3| = |z_2 - z_3|$. Now,consider the vector $\vec{BA} = z_1 - z_2$ and $\vec{BC} = z_3 - z_2$. In $	riangle ABC$,$\angle B = 45^{\circ}$ and $AC = BC$. Using the rotation formula,$\frac{z_1 - z_2}{z_3 - z_2} = \sqrt{2} e^{i\pi/4} = \sqrt{2}(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}) = \sqrt{2}(\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}) = 1 + i$. Similarly,$\frac{z_2 - z_1}{z_3 - z_1} = \sqrt{2} e^{-i\pi/4} = 1 - i$. Multiplying these two,we get $\frac{(z_1 - z_2)(z_2 - z_1)}{(z_3 - z_2)(z_3 - z_1)} = (1+i)(1-i) = 1 - i^2 = 2$. $-(z_1 - z_2)^2 = 2(z_3 - z_2)(z_3 - z_1)$. Since $z_3 - z_1 = -(z_1 - z_3)$,we have $-(z_1 - z_2)^2 = 2(z_3 - z_2)(-(z_1 - z_3))$. Therefore,$(z_1 - z_2)^2 = 2(z_1 - z_3)(z_3 - z_2)$.

If the vertices $A, B$ and $C$ of an isosceles $\triangle ABC$ are respectively $z_1, z_2$ and $z_3$ and if $\angle C=90^{\circ}$,then

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