question_answer

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

Answer:

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

Draw \[OM\bot AB\], and join OA and OB.

In \[\Delta \text{ }OAM\]and \[\Delta \text{ O}MB\],

\[OA=OB\]                       (Radii)

\[OM=OM\]           (Common)

\[\angle OMA=\angle OMB\]                 (Each \[90{}^\circ \])

\[\therefore \]                  \[\Delta \,OAM\cong \Delta \,OMB\]          (R.H.S. cong.)

\[\therefore \]                  \[\angle OAM=\angle OBM\]                    (CECT)

Also, \[\angle OAP=\angle OBR=90{}^\circ \] (Line joining point of contact of tangent to centre is perpendicular on 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