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Let $S$ be a circle centered at the origin in the Argand Plane. Let A($z_1$), B($z_2$) and C($z_3$) be 3 distinct points on the circle $S$. In $\Delta ABC$, let the altitude from A on BC be extended to cut the circle $S$ at the point D$(z_4)$ . Then show:- $$ z_4 = -\frac{z_2\cdot z_3}{z_1} $$

My attempt: I tried to use the fact that AD is perpendicular to BC, so by complex slope: $\frac{z_4-z_1}{\bar{z_4} - \bar{z_1}} + \frac{z_3-z_2}{\bar{z_3} - \bar{z_2}} = 0 $, but this just rearranges to get $(z_4 - z_1) \odot (z_3 - z_2) = 0$ where $a\odot b$ represents the dot product of $a$ and $b$.

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Without restriction, the circle is the unit circle. Write $z_k=\exp(it_k)$, $1\le k\le 4$, $0\le t_k\le 2\pi$. Let us assume $z_1,z_2,z_4,z_3$ are coming in this cyclic order. Then the measure of the angle between the line $z_1z_4$ and the line $z_2z_3$ is half the sum of the measures of the arcs $z_1z_2$ and $z_4z_3$. The condition is thus $$ \frac 12\Big(\ (t_2-t_1)+(t_3-t_4)\ \Big) = \frac \pi2\ . $$ Multiply with $2$ and apply $x\to\exp(ix)$ on both sides to get $$ \frac {z_2z_3}{z_1z_4}=-1\ . $$ (The relation is symmetric w.r.t. the two pairs, $(z_1,z_4)$ and $(z_2,z_3)$, so the ordering is not important, switch $x_2, x_3$ if the oder order is needed.)

Later edit: Here, we have used the following property of an "interior angle" delimited by two chords in a circle. Let $X$ be the intersection of the two chords $AC$ and $BD$. Then the marked red angle in $X$ is exterior w.r.t. the triangle $\Delta AXD$, so its measure is the sum of the two green marked angles in $A$ and $D$. These two angles are inscribed in the circle, their measure is each half of the measure of the arc delimited on the circle, $\overset\frown{CD}$ (for $\hat A$) and $\overset\frown{AB}$ (for $\hat D$)...

Interior angle in a circle

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  • $\begingroup$ I am not able to see why the measure of the angle between the line $z_1z_4$ and the line $z_2z_3$ is half the sum of the measures of the arcs $z_1z_2$ and $z_4z_3$. Could you elaborate on this bit? $\endgroup$ – Kaind Jun 25 '18 at 17:51
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Besides assuming that it's the unit circle, note that rotating the circle (i.e. multiplying everything by $e^{i\theta}$ for some $\theta$) makes no difference. Since $AD$ and $BC$ are perpendicular, we can assume that $z_1$ and $z_4$ have the same $x$ value while $z_2$ and $z_3$ have the same $y$ value. Then for some $\alpha$ we have $z_1 = e^{i\alpha}, z_4 = e^{-i\alpha}$, so $z_1z_4 = 1$; and for some $\beta$ we have $z_2 = -e^{i\beta}, z_3 = e^{-i\beta}$, so $z_2z_3 = -1$.

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  • $\begingroup$ The triangle ABC is given to you, so once you fix $\alpha$ , the other angle is automatically fixed; hence you cannot choose some $\beta$ so that $z_2z_3 = -1$. You have shown that $\exists$ a triangle ABC, such that $z_1z_4 = 1$ and $z_2z_3 = -1 $. $\endgroup$ – Kaind Jun 25 '18 at 17:52

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