Question

Prove that the tangents drawn at the end points of a chord of a circle make equal angles with the chord.

Answer 1

Given a circle of radius OA centred at O with chord AB and tangents PQ & RS are drawn from point A and B respectively.



Draw OM $\perp$ AB, and join OA and OB.

In $△$ OAM and $△$OMB,.

OA = OB$\left(\text{Radii}\right)$

OM = OM$\left(\text{Common}\right)$

$\angle$ OMA =$\angle$OMB $\left(Each{90}^{\circ }\right)$

$\therefore △$ OAM = $\cong △$ OBM $(R.H.S.Cong)$
$\therefore \angle$ OAM =$\angle$ OBM $\left(CPCT\right)$
Also, $\angle$ OAP = $\angle$ OBR = ${90}^{\circ }(\because \text{Line joining point of contact of tangent to centre is perpendicular to it})$

On addition,

$\angle$ OAM + $\angle$ OAP = $\angle$ OBM +$\angle$ OBR

$\Rightarrow \angle$ PAB = $\angle$RBA

$\Rightarrow \angle$ PAQ - $\angle$ PAB = $\angle$RBS - $\angle RBA$

$\Rightarrow \angle$ QAB = $\angle$ SBA
Hence Proved.