If the chord $A B$ of a circle is parallel to the tangent at $C .$ Then prove that $A C = B C .$

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Solution

$A B$ is a chord of the circle, and by taking a point $C$ on the circle . A tangent is drawn
such that line $A B$ is parallel to the point $C$ .
Now, Join $A C$ and $B C$ .
Now Let $O$ be the center of the circle
Draw $O M$ perpendicular to $A B$ . and join $O$ and $C$
$\begin{matrix} A s, ∠ O C N = 90^{\circ} a n d ∠ O C S = 90^{\circ} A l s o A M = M B N o w, ∠ B M C + ∠ N C O = 180^{\circ} s o, p o i n t s, M, O a n d C l i e i n a l i n e . I n Δ C M A a n d Δ C M B C M = C M [c o m m o n] ∠ C M A = ∠ C M B = 90^{\circ} . a n d, A M = M B s o, Δ C M A ≃ Δ C M B [b y S A S c o n g r u e n t .] H e n c e, A C = B C p o v e d . \end{matrix}$

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