1.

DatePaga Not chausn at thsve that the fansennid-pantapato the chrdjaning thePank

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 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



Discussion

No Comment Found