Triangle ABC is isosceles in which AB=AC circumscribed about a circle. Prove that base is bisected by the point of contact.
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let base be bisected at x ,then,ap=aq(tangents).
AB=AC,AP=AQ THEREFORE,AB-AP=AC-AQ.
That is PB=CQ----1
But BP=BX and QC=CX (tangets) -----2
Fom 1 and 2,we prove BX=CX
AB=AC,AP=AQ THEREFORE,AB-AP=AC-AQ.
That is PB=CQ----1
But BP=BX and QC=CX (tangets) -----2
Fom 1 and 2,we prove BX=CX
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Answer:
Given that,
AB=AC
Step-by-step explanation:
there are 3 pairs of tangents
AR=AQ-------------1
BR=BP--------------2
CQ=CP-------------3
AB=AC [Given]
AR+BQ=AQ+CQ
AR+BQ=AR+CQ [From 1]
BQ=CQ
BP=CP [From 2 & 3]
Hence, proved that point of contact of the base(P) bisects BC.
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