Q. 265.0( 2 Votes )

In Fig. 4, an iso

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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