Prove that the length of the tangents drawn from an external point to a circle are equal.
[ JKBOSE- 2019, 18, 17, 14, most imp. ]
Given a circle with centre 0.
To prove $$\longrightarrow A C=B C$$
Proof : Since the tangent drawn to a circle make 90 degree at the point of contact.
In $$\triangle A O C$$ and $$\triangle B O C$$
$$\angle O A C=\angle C B O=90^{\circ}$$ each
$$\quad O C=O C$$ (Common)
$$AO=O B$$ (radii of the same circle)
$$\triangle AOC \sim \triangle B O C($$ by $$R.H.S)$$
$$BY CP C T, A C=C B$$
R.H.S Stand for Right, Hypotenuse, Side
Hence proved
