Question

drag and drop an answer to each box to correctly complete the proof.
it is given that quadrilateral defg is inscribed in circle p. because a circle measures 360°, ( moverparen{efg}+moverparen{gde}=360^{circ} ). by the, ( \frac{1}{2}moverparen{efg}+\frac{1}{2}moverparen{gde}=180^{circ} ).
by the inscribed angles theorem, ( =\frac{1}{2}moverparen{gde} ) and ( =\frac{1}{2}moverparen{efg} ). this means ( mangle d + mangle f = 180^{circ} ) by the

Explanation:

Step1: Apply the division property of equality

Given \(m\overarc{EFG}+m\overarc{GDE} = 360^{\circ}\). Divide both sides of the equation by \(2\). Using the division property of equality (\(a = b\Rightarrow\frac{a}{c}=\frac{b}{c},c
eq0\)), we get \(\frac{1}{2}m\overarc{EFG}+\frac{1}{2}m\overarc{GDE}=180^{\circ}\).

Step2: Use the inscribed - angle theorem

The inscribed - angle theorem states that an inscribed angle in a circle is half the measure of its intercepted arc. So, \(m\angle D=\frac{1}{2}m\overarc{GDE}\) (angle \(D\) intercepts arc \(GDE\)) and \(m\angle F=\frac{1}{2}m\overarc{EFG}\) (angle \(F\) intercepts arc \(EFG\)).

Step3: Substitute using the substitution property

Substitute \(m\angle D\) for \(\frac{1}{2}m\overarc{GDE}\) and \(m\angle F\) for \(\frac{1}{2}m\overarc{EFG}\) in the equation \(\frac{1}{2}m\overarc{EFG}+\frac{1}{2}m\overarc{GDE}=180^{\circ}\). By the substitution property (if \(a = b\) and \(c = d\), then \(a + c=b + d\)), we have \(m\angle D+m\angle F = 180^{\circ}\).

Answer:

First blank: division property of equality; Second blank: \(m\angle D\); Third blank: \(m\angle F\); Fourth blank: substitution property.