In the figure, B is a point outside circle O, and tangents BA and BC are drawn. BO cuts chord AC at D. Give a…

Question

In the figure, B is a point outside circle O, and tangents BA and BC are drawn. BO cuts chord AC at D. Give a reason why OCB = 90°?

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Here are the solutions to the geometry problem:

2.4.1 Give a reason why $\angle OCB = 90^\circ$ .

Identify the relationship between OC and BC.
OC is a radius of the circle, and BC is a tangent to the circle at point C.
Apply the tangent-radius theorem.
A tangent to a circle is perpendicular to the radius at the point of tangency.
$\boxed{\text{Radius OC is perpendicular to tangent BC at C.}}$
2.4.2 Prove with reasons that OABC a cyclic quadrilateral?
Identify known angles in quadrilateral OABC.
From 2.4.1, $\angle OCB = 90^\circ$ (Radius $\perp$ tangent).
Similarly, OA is a radius and BA is a tangent at A, so $\angle OAB = 90^\circ$ (Radius $\perp$ tangent).
Check for the cyclic quadrilateral condition.
The sum of opposite angles $\angle OAB$ and $\angle OCB$ is $90^\circ + 90^\circ = 180^\circ$ .
$\angle OAB + \angle OCB = 90^\circ + 90^\circ = 180^\circ$
Conclude that OABC is a cyclic quadrilateral.
A quadrilateral is cyclic if the sum of a pair of opposite angles is $180^\circ$ .
$\boxed{\text{OABC is a cyclic quadrilateral because opposite angles } \angle OAB \text{ and } \angle OCB \text{ sum to } 180^\circ.}$
2.4.3 Let $A_2 = y$ . Name the other THREE angles equal to $y$ , with reasons.
Given $A_2 = \angle OAC = y$ .
Use properties of $\triangle OAC$ .
OA and OC are radii of the circle, so $OA = OC$ . Therefore, $\triangle OAC$ is an isosceles triangle.
$\angle OCA = \angle OAC = y$
Reason: Angles opposite equal sides in an isosceles triangle.
Use properties of cyclic quadrilateral OABC.
Since OABC is a cyclic quadrilateral (proven in 2.4.2), angles subtended by the same arc are equal.
$\angle OBC$ and $\angle OAC$ are subtended by arc OC.
$\angle OBC = \angle OAC = y$
Reason: Angles in the same segment (subtended by arc OC in cyclic quad OABC).
$\angle OBA$ and $\angle OCA$ are subtended by arc OA.
$\angle OBA = \angle OCA = y$
Reason: Angles in the same segment (subtended by arc OA in cyclic quad OABC).
$\boxed{\text{The three angles equal to } y \text{ are:}}$ $\boxed{\text{1. } \angle OCA \text{ (Angles opposite equal sides in isosceles } \triangle OAC)}$ $\boxed{\text{2. } \angle OBC \text{ (Angles in the same segment of cyclic quad OABC, subtended by arc OC)}}$ $\boxed{\text{3. } \angle OBA \text{ (Angles in the same segment of cyclic quad OABC, subtended by arc OA)}}$
2.4.4 Explain why AB is a tangent to the circle through A, O and D.
Identify known angles related to the line AB and the circle through A, O, D.
We know that $\angle OAB = 90^\circ$ because OA is a radius of the original circle and AB is a tangent to the original circle at A (Radius $\perp$ tangent).
Determine the angle subtended by chord AO in the circle through A, O, D.
In $\triangle OAC$ , $OA=OC$ (radii), so $\triangle OAC$ is isosceles.
The line BO passes through D and bisects $\angle AOC$ (due to $\triangle OAB \cong \triangle OCB$ ). In an isosceles triangle, the angle bisector from the vertex angle is also the altitude to the base.
Therefore, $OD \perp AC$ , which means $\angle ODA = 90^\circ$ .
Apply the converse of the tangent-chord theorem.
Consider the circle passing through A, O, and D. The line AB forms an angle $\angle OAB$ with the chord AO. The angle subtended by the chord AO in the alternate segment of this circle is $\angle ODA$ .
Since $\angle OAB = 90^\circ$ and $\angle ODA = 90^\circ$ , we have $\angle OAB = \angle ODA$ .
$\boxed{\text{AB is a tangent to the circle through A, O, D because the angle between the line AB and chord AO (i.e., } \angle OAB) \text{ is equal to the angle subtended by the chord AO in the alternate segment (i.e., } \angle ODA).}$ $\boxed{\text{Both angles are } 90^\circ \text{ (Converse of the tangent-chord theorem).}}$

In the figure, B is a point outside circle O, and tangents BA and BC are drawn. BO cuts chord AC at D. Give a reason why OCB = 90°?

Step-by-step solution

Identify the relationship between OC and BC.

Apply the tangent-radius theorem.

Identify known angles in quadrilateral OABC.

Check for the cyclic quadrilateral condition.

Conclude that OABC is a cyclic quadrilateral.

Use properties of $\triangle OAC$ .

Use properties of cyclic quadrilateral OABC.

Identify known angles related to the line AB and the circle through A, O, D.

Determine the angle subtended by chord AO in the circle through A, O, D.

Apply the converse of the tangent-chord theorem.

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More Mathematics Questions

In the figure, B is a point outside circle O, and tangents BA and BC are drawn. BO cuts chord AC at D. Give a reason why OCB = 90°?

Step-by-step solution

Identify the relationship between OC and BC.

Apply the tangent-radius theorem.

Identify known angles in quadrilateral OABC.

Check for the cyclic quadrilateral condition.

Conclude that OABC is a cyclic quadrilateral.

Use properties of △OAC\triangle OAC△OAC.

Use properties of cyclic quadrilateral OABC.

Identify known angles related to the line AB and the circle through A, O, D.

Determine the angle subtended by chord AO in the circle through A, O, D.

Apply the converse of the tangent-chord theorem.

Still stuck on this one?

More Mathematics Questions

Use properties of $\triangle OAC$ .