Question

given: quadrilateral abcd is inscribed in circle f, and $overline{ad}$ is a diameter of circle f. prove: the diagonal $overline{bd}$ of quadrilateral abcd forms a right - angle, $angle abd$, with side $overline{ab}$. move options to the blanks to complete the proof of this property. because $overline{ad}$ is a diameter of circle f, $mwidehat{aed}=180^{circ}$. then, by the inscribed angle theorem, $mwidehat{aed}=square$. by substitution, this means that $180^{circ}=square$. by division, $90^{circ}=square$. $\frac{1}{2}mangle abd$, $mangle abd$, $2cdot mangle abd$, $mangle afd$, $2cdot mangle afd$

Explanation:

Step1: Apply Inscribed - Angle Theorem

The Inscribed - Angle Theorem states that the measure of an inscribed angle is half the measure of its intercepted arc. Here, $\angle ABD$ is an inscribed angle and $\overset{\frown}{AED}$ is the intercepted arc. So, $m\overset{\frown}{AED}=2\cdot m\angle ABD$.

Step2: Substitute the value of $m\overset{\frown}{AED}$

We know that $m\overset{\frown}{AED} = 180^{\circ}$. Substituting this into the equation from Step 1, we get $180^{\circ}=2\cdot m\angle ABD$.

Step3: Solve for $m\angle ABD$

Dividing both sides of the equation $180^{\circ}=2\cdot m\angle ABD$ by 2, we have $90^{\circ}=m\angle ABD$.

Answer:

First blank: $2\cdot m\angle ABD$
Second blank: $2\cdot m\angle ABD$
Third blank: $m\angle ABD$