Question

please fast 50 points

Answer 1

we know that tangents from an exterior point to a circle are always equal,
AB = AC

AB = AC = r (given)

OB = OC = r (radius of same circle)

AB = BO = OC = CA

Both radius make 90° with tangents,
angle ABO = angle ACO = 90°

It shows that it is a square

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Answer 2

in ∆ABO and ∆ACO,
angle ABO=angle ACO=90° [ because tangent and radius make 90° at point of contact ]
AO=AO [common]
BO= CO [Radii of same circle]
So, by RHS congruence rule
∆ABO congruent to ∆ACO

so, AB= AC = r -(1) [ by CPCT]
and BO=CO = r -(2)
from (1) and (2) ,
AB=AC=BO=CO= r
and angle ABO = ACO = 90°
so quad. ABOC is a square


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