Question

identifying properties of angles intercepting the same arc
use the diagram to complete the statements.
the measure of angle ejb is □ the measure of angle boe.
the measure of angle bde is □ the measure of angle boe.
the measure of angle oed is □ the measure of angle obd.

Explanation:

To solve this, we use properties of circles (inscribed angles, central angles, and tangent - radius properties):

For the first statement (∠EJB and ∠BOE)

• ∠EJB is an inscribed angle that intercepts arc EB. ∠BOE is a central angle that also intercepts arc EB.
• The measure of an inscribed angle is half the measure of its intercepted central angle. So, \(m\angle EJB=\frac{1}{2}m\angle BOE\). Thus, the measure of angle EJB is half the measure of angle BOE.

For the second statement (∠BDE and ∠BOE)

• EB is a chord, and ED is a tangent (since ED is a line touching the circle at E, and OE is a radius, so \(OE\perp ED\)). Also, BD is a secant.
• The measure of an angle formed by a tangent and a secant outside the circle is half the difference of the measures of the intercepted arcs. But in this case, we can also use the fact that ∠BDE and ∠EJB are related (since \(JB\parallel ED\) maybe? Or by triangle similarity and angle relations). But more directly, since ∠EJB is half of ∠BOE, and ∠BDE is equal to ∠EJB (alternate interior angles if \(JB\parallel ED\)) or by the tangent - chord angle theorem. Wait, actually, ∠BDE is an angle formed by tangent ED and secant BD. The measure of ∠BDE is half the measure of the intercepted arc BE (the minor arc). And ∠BOE is the central angle for arc BE. So, \(m\angle BDE = \frac{1}{2}m\angle BOE\). So the measure of angle BDE is half the measure of angle BOE.

For the third statement (∠OED and ∠OBD)

• OE and OB are radii of the circle, so \(OE = OB\). Also, \(OE\perp ED\) (tangent - radius property), so ∠OED = 90°.
• Triangle OBD: OB is a radius, OD is a line from the center to D. Also, since ED is tangent at E and BD is a secant, we can show that triangles OED and OBD are right - angled (∠OED = 90° and ∠OBD = 90°? Wait, no. Wait, OB is a radius, and BD is a tangent? Wait, no, the diagram shows BD as a secant? Wait, maybe BD is tangent at B? Wait, the diagram: E and B are points on the circle, O is the center. ED is tangent at E, BD is a line from D to B. If we assume that BD is tangent at B, then \(OB\perp BD\), so ∠OBD = 90°. And \(OE\perp ED\), so ∠OED = 90°. So ∠OED is equal to ∠OBD.

Final Answers

1. The measure of angle EJB is \(\boldsymbol{\text{half}}\) the measure of angle BOE.
2. The measure of angle BDE is \(\boldsymbol{\text{half}}\) the measure of angle BOE.
3. The measure of angle OED is \(\boldsymbol{\text{equal to}}\) the measure of angle OBD.

Answer:

To solve this, we use properties of circles (inscribed angles, central angles, and tangent - radius properties):

For the first statement (∠EJB and ∠BOE)

• ∠EJB is an inscribed angle that intercepts arc EB. ∠BOE is a central angle that also intercepts arc EB.
• The measure of an inscribed angle is half the measure of its intercepted central angle. So, \(m\angle EJB=\frac{1}{2}m\angle BOE\). Thus, the measure of angle EJB is half the measure of angle BOE.

For the second statement (∠BDE and ∠BOE)

• EB is a chord, and ED is a tangent (since ED is a line touching the circle at E, and OE is a radius, so \(OE\perp ED\)). Also, BD is a secant.
• The measure of an angle formed by a tangent and a secant outside the circle is half the difference of the measures of the intercepted arcs. But in this case, we can also use the fact that ∠BDE and ∠EJB are related (since \(JB\parallel ED\) maybe? Or by triangle similarity and angle relations). But more directly, since ∠EJB is half of ∠BOE, and ∠BDE is equal to ∠EJB (alternate interior angles if \(JB\parallel ED\)) or by the tangent - chord angle theorem. Wait, actually, ∠BDE is an angle formed by tangent ED and secant BD. The measure of ∠BDE is half the measure of the intercepted arc BE (the minor arc). And ∠BOE is the central angle for arc BE. So, \(m\angle BDE = \frac{1}{2}m\angle BOE\). So the measure of angle BDE is half the measure of angle BOE.

For the third statement (∠OED and ∠OBD)

• OE and OB are radii of the circle, so \(OE = OB\). Also, \(OE\perp ED\) (tangent - radius property), so ∠OED = 90°.
• Triangle OBD: OB is a radius, OD is a line from the center to D. Also, since ED is tangent at E and BD is a secant, we can show that triangles OED and OBD are right - angled (∠OED = 90° and ∠OBD = 90°? Wait, no. Wait, OB is a radius, and BD is a tangent? Wait, no, the diagram shows BD as a secant? Wait, maybe BD is tangent at B? Wait, the diagram: E and B are points on the circle, O is the center. ED is tangent at E, BD is a line from D to B. If we assume that BD is tangent at B, then \(OB\perp BD\), so ∠OBD = 90°. And \(OE\perp ED\), so ∠OED = 90°. So ∠OED is equal to ∠OBD.

Final Answers

1. The measure of angle EJB is \(\boldsymbol{\text{half}}\) the measure of angle BOE.
2. The measure of angle BDE is \(\boldsymbol{\text{half}}\) the measure of angle BOE.
3. The measure of angle OED is \(\boldsymbol{\text{equal to}}\) the measure of angle OBD.