Answer :
We have figure,
Given: AB = AC and ∆ABC circumscribe a circle with the circle touching the side AB at R, AC at Q and BC at P.
To Prove: P bisects the side of the triangle, BC.
⇒ BP = PC
Proof: ∵, RB and BP are tangents to the circle from the same point. ⇒ BP = BR …(i)
QC and CP are tangents to the circle from the same point. ⇒ QC = CP …(ii)
Also, RA and AQ are tangents to the circle from the same point. ⇒ AQ = AR …(iii)
Adding equations (i), (ii) and (iii), we get
BP + QC + AQ = BR + CP + AR
⇒ (AQ + QC) + BP = (AR + BR) + CP
⇒ AC + BP = AB + CP
⇒ BP = CP [∵, BP = CP]
Or
Given: A chord AB of the larger of the two concentric circles having mid-point O, touching the smaller circle at point C.
To prove: AC = AB
Construction: Join O to C.
Proof: AB is tangent to the inner circle and OC is meeting AB at C.
⇒ ∠OCB = ∠OCA = 90°, which means that OC is perpendicular to AB.
AB is also the chord of the outer circle, on which OC is perpendicular to.
∵, According to the chord theorem, the perpendicular from the centre of a circle to a chord bisects the chord.
Thus, this only justifies that OC bisects AB at the point of its contact.
⇒ AC = CB
Hence, proved.
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