prove that the tangent drawn at the end of an arc of a circle is paral

1.	prove that the tangent drawn at the end of an arc of a circle is parallel to the chord joining the end points of the arc
Answer» 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 ⟂PT [Radius is ⟂ to a tangent through the point of contact] ∠OPT = 90° Since P is the midpoint of Arc APB Arc AAP = 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, AD\|\|PT

prove that the tangent drawn at the end of an arc of a circle is parallel to the chord joining the end points of the arc

Answer»

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 ⟂PT

[Radius is ⟂ to a tangent through the point of contact]

∠OPT = 90°

Since P is the midpoint of Arc APB

Arc AAP = 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, AD||PT