Question

Prove that the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc.

Answer 1

To prove :: $AB||PT$

Construction: join $OA,OB,\&OP$

Proof:

$\Rightarrow$$OP⟂PT$ [Radius is ⟂ to tangent through a point of contact]

$\Rightarrow$$\angle OPT=90°$

Since P is the midpoint of Arc $APB$

$\Rightarrow$$Arc\left(AAP\right)=arc\left(BP\right)$

$\Rightarrow$$\angle AOP=\angle BOP$

$\Rightarrow$$\angle AOM=\angle BOM$

$\Rightarrow$In $\Delta AOM\&\Delta BOM$

$\Rightarrow$$OA=OB=r$

$\Rightarrow$$OM=OM$ (Common)

$\Rightarrow$$\angle AOM=\angle BOM$ (proved above)

$\Rightarrow$$\angle AOM\cong \angle BOM$ (by $SAS$ congruence axiom)

$\Rightarrow$$\angle AMO=\angle BMO$ $(C.P.C.T)$

$\Rightarrow$$\angle AMO+\angle BMO=180°$

$\Rightarrow$$\angle AMO=\angle BMO=90°$

$\Rightarrow$$\angle BMO=\angle OPT=90°$

But, they are corresponding angles. Hence, $AB||PT$