Question

Two tangents PT and PT' are drawn to a circle, with centre O, from an external point P. Prove that $\angle$TPT'$=2\angle$OTT'.

Answer 1

Given: A circle with centre $O$ and $PT,P{T}^{\prime }$ are tangents on the circle from the point $P$ outside it.

To prove :- $\angle TP{T}^{\prime }=2\angle DT{T}^{\prime }$

Proof:- Let $\angle TP{T}^{\prime }=x^0$

we know that the tangents to a circle from an external point are equal. So, $PT=P{T}^{\prime }$

Since, the angles opposite to the equal sides of a triangle are equal. So,

$PT=P{T}^{\prime }\Rightarrow \angle PT{T}^{\prime }+\angle P{T}^{\prime }T=\angle PT{T}^{\prime }$

Also, Sum of the angles of a $△$ is ${180}^o$

$\therefore \angle TP{T}^{\prime }+\angle PT{T}^{\prime }+\angle P{T}^{\prime }T={180}^o$

$\Rightarrow x^o+2\angle PT{T}^{\prime }={180}^o$ $[\because \angle P{T}^{\prime }T=\angle PT{T}^{\prime }]$

$\Rightarrow \angle PT{T}^{\prime }=\frac{1}{2}({180}^o-x)=({90}^o-\frac{1}{2}{x}^o)$

But $PT$ is a tangent and $OT$ is the radius of the given circle.

$\therefore \angle OT{T}^{\prime }+\angle PT{T}^{\prime }={90}^o$

$\Rightarrow \angle OT{T}^{\prime }={90}^o-({90}^o-\frac{1}{2}{x}^o)$

$\Rightarrow \angle OT{T}^{\prime }=\frac{1}{2}{x}^o$

$\Rightarrow \angle OT{T}^{\prime }=\frac{1}{2}\angle TP{T}^{\prime }$

$\Rightarrow \angle TP{T}^{\prime }=2\angle OT{T}^{\prime }$

Hence proved