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liberstina [14]
3 years ago
8

Circle geometry help?

Mathematics
1 answer:
Tanya [424]3 years ago
4 0
Question (i):

In ABOC, ∠OCA = 90° (tangent and the radius meet at perpendicular angle)
∠ABO = 90° (same as above)
∵ ∠OCA + ∠ABO = 180°, ABOC is a cyclic quadrilateral

In AOGC, ∠OCA = 90° (proven)
∠OGA is also 90°<span> (line from centre bisects the chord of a circle at a right angle)
</span>∵∠OCA = ∠OGA, AOGC is a cyclic quadrilateral (equal angles standing on the same arc on a cyclic quad)

Question (ii):
Since ABOC is a cyclic quad, ∠OGF = ∠OAC (exterior angle of a cyclic quadrilateral is equal to the interior opposite angle)

Question (iii):
Let ∠OGF = α
From (ii), ∠OAC = α

AC = AB <span>(tangents meeting at an external point are equal)
AO is common
OC = OB (radii)

</span>∠BAO = α (corresponding angles in congruent triangles)
∠BOC = π - 2α <span>(supplementary angles; opposite angles of a cyclic quad are supplementary)
</span>
It should be pretty simple after this, I'll finish it in a little bit.
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