Question

Prove : length of 2 tangents drawn from an external point to circle are equal​s.​

Answer 1

Dɪᴀɢʀᴀᴍ :

$\setlength{\unitlength}{22} \begin{picture}(6,6) \linethickness{0.6} \put(2,2){\oval(2,2)}\put(2,2){\circle*{0.1}}\qbezier(2,2)(1.5,2.9)(1.5,2.9)\qbezier(2,2)(1.5,1)(1.5,1.1)\qbezier(3.5,3.5)(3,3.4)( - 1,2)\put(2,2){\line( - 1,0){0.5}}\qbezier( - 1,2)(3,0.6)(3.5,0.4)\put(1,2){\line( - 1,0){0.5}}\put(0,2){\line( - 1,0){0.5}}\put( - 0.6,2){\line( - 1,0){0.3}}\qbezier(1.3,2.8)(1.4,2.2)(1.7,2.5)\qbezier(1.3,1.2)(1.3,1.6)(1.8,1.5)\put(2.05,2){ \bf O }\put(1.4,3){ \bf A }\put(1.4,0.7){ \bf C }\put( - 1.5,1.9){ \bf B }\end{picture}$

Gɪᴠᴇɴ :

• $\footnotesize\textsf{ AB and BC are two tangents meeting to external point }$

Tᴏ Pʀᴏᴠᴇ :

• $\footnotesize\sf \overline{AB} \: =\: \overline{BC}$

Tᴏ Pʀᴏᴏғ :

$\footnotesize\sf In \:\triangle AOB \: \& \: \triangle BOC$

$\footnotesize\mapsto\:\:\sf AB^{2}\: +\: AO^{2}\: =\: OB^{2}.......(1)$

$\footnotesize\mapsto\:\:\sf BC^{2}\: +\: CO^{2}\: =\: OB^{2}.......(2)$

$\footnotesize\rm\underline{From \:Equation\:(1) \:\&\:(2)}$

$\footnotesize\hookrightarrow\:\:\sf AB^{2}\: +\: AO^{2}\: =\: BC^{2}\: +\: CO^{2}$

$\footnotesize\hookrightarrow\:\:\sf AO \:= \:CO.....(radius\: of \:circle)$

$\footnotesize\hookrightarrow\:\:\sf AB^{2}\: +\: \cancel{CO^{2}} \: =\: BC^{2}\: +\: \cancel{CO^{2}}$

$\footnotesize\hookrightarrow\:\:\sf AB^{2}\: =\: BC^{2}$

$\footnotesize\hookrightarrow\:\:\sf \overline{AB} \: =\: \overline{BC}$

$\footnotesize\star\:\bf\underline{Hence\: Proved}$