Question

Prove that the intercept of a tangent between two parallel tangents to a circle subtends a right angle at a center.

Answer 1

R.E.F image

Prove that $\angle BOA={90}^o$.

In $\Delta P_{1}AO$ and $\Delta CAO$

$AO=AO$ (common)

$O{P}_1=OC$ (Radius)

$A{P}_1=AC$ (tangent from external point)

$\Delta P_{1}AP\cong \Delta CAO$ (sss criterion)

Similarly, $\Delta P_{2}BO\cong \Delta CBO$

$\angle P_{1}OA+\angle AOC+\angle COB+\angle P_{2}OB={180}^o$ (liner pair).

$\angle P_{1}OA=\angle AOC$ and $\angle P_{2}OB=\angle COB$

$2(\angle AOC+\angle COB)={180}^o$

$\angle AOC+\angle COB={90}^o$

$\angle BOA={90}^o$

Hence intercept of a tangent between two parallel tangent to a circle

subtends a right angle at a center.