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

Answer:

Construction: join OA ,OB, & OP

Proof: OP ⟂PT

[Radius is ⟂ to  tangent through 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

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Answer 2

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  tangent through 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,proved

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