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

Given: A circle with Centre O, P is the midpoint of Arc APB. PT is a tangent to the circle at P.

To Prove: AB || PT

Construction: Join OA ,OB, & OP

Proof: $OP\perp PT$ [Radius is $\perp$ to tangent through point of contact]

∠OPT= 90°

Since P is the midpoint of Arc APB

Arc AP = arc BP
∠AOP = ∠BOP
∠AOM= ∠BOM

In ∆ AOM & ∆BOM

OA= OB= r
OM = OM (Common)
∠AOM= ∠BOM (proved above)
∠AOM≅∠BOM (by SAS congruency axiom)
∠AMO = ∠BMO (c.p.c.t)
∠AMO + ∠BMO= 180°
∠AMO = ∠BMO= 90°
∠BMO = ∠OPT= 90°

But they are corresponding angles. Hence, AP || PT