<s>[INST] Prove the point is the midpoint of a segment Let a circle (O) and a point A outside of the circle. AB, AC are tangents of (O) (B,C $\in$ (O)). BD is an diameter of (O). CK perpendicular to BD (K $\in$ BD). Let I is intersection of CK and AD. Prove that I is midpoint of KC
 [/INST] 
Without any loss of generality we can assume $B(a,0),O(0,0)$ so that $D(-a,0)$ the equation of the circle $(x-0)^2+(y-0)^2=(a-0)^2\implies x^2+y^2=a^2$
Using this, the equation of the tangent at any point $P(a\cos\theta,a\sin\theta)$ will be
$xa\cos \theta+ya\sin\theta=a^2\implies x\cos \theta+y\sin \theta=a$
So, $AB: x=a--->(1)$ (putting $\theta=0$)
Let $C$ be $(a\cos \beta,a\sin \beta)$ so $CA: x\cos\beta+y\sin\beta=a--->(2)$ (putting $\theta=\beta$)
Solving for $y$ we get, $y=\frac{a(1-\cos\beta)}{\sin \beta}$ so $A(a,\frac{a(1-\cos\beta)}{\sin \beta})$
$AD$ will be $$\frac{y-0}{x+a}=\frac{0-\frac{a(1-\cos\beta)}{\sin \beta})}{-a-a}=\frac{1-\cos\beta}{2\sin \beta}\implies 2y=(x+a)\frac{1-\cos\beta}{\sin \beta}$$
Clearly, the abscissa of $I$ will be $a\cos \beta$
So, $2y=(a\cos \beta+a)\frac{1-\cos\beta}{\sin \beta}=a\sin\beta\implies y=\frac{a\sin\beta}2$
Hence, $I(a\cos\beta,\frac{a\sin\beta}2)$ is the midpoint of $C(a\cos\beta,a\sin\beta)$ and $K(a\cos\beta,0)$
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