Question

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

Answer 1

[FIGURE IS IN THE ATTACHMENT]

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


HOPE THIS WILL HELP YOU... .

Answer 2

Answer:

To prove :: AB∣∣PT

Construction: join OA,OB,&OP

Proof:

⇒OP⊥PT [Radius is ⟂ to tangent through a 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 congruence axiom)

⇒∠AMO=∠BMO (C.P.C.T)

⇒∠AMO+∠BMO=180°

⇒∠AMO=∠BMO=90°

⇒∠BMO=∠OPT=90°

But, they are corresponding angles. Hence, AB∣∣PT

solution