Question

complete the proof for the second case of the inscribed angle theorem to show that m∠dba = 1/2mda⌢. given: ∠dba is inscribed in circle c, with diameter bx⌢. prove: m∠dba = 1/2mda⌢. proof: let m∠abx = x° and m∠dbx = y°. drag the choices into the boxes to form a correct proof. show hints draw ac and dc. △abc and △dbc are isosceles triangles because radii of a circle are congruent, so ac = bc = dc. then m∠bac = z° and m∠bdc = m∠acx = 2x° and m∠dcx = by the isosceles triangle theorem. by the exterior angle theorem.

Explanation:

Step1: Recall circle - related properties

Since \(BX\) is a diameter of circle \(C\), and \(\triangle ABC\) and \(\triangle DBC\) are isosceles triangles because \(AC = BC=DC\) (radii of the circle). Let \(m\angle ABX=x^{\circ}\) and \(m\angle DBX = y^{\circ}\), so \(m\angle DBA=x + y\).

Step2: Use the exterior - angle theorem

The measure of an exterior angle of a triangle is equal to the sum of the measures of the two non - adjacent interior angles. In \(\triangle ABC\), if \(m\angle BAC = z^{\circ}\), and in \(\triangle DBC\), if \(m\angle BDC=z^{\circ}\) (base - angles of isosceles triangles \(\triangle ABC\) and \(\triangle DBC\)). Let the central angle corresponding to arc \(DA\) be \(m\overset{\frown}{DA}=2(x + y)\).

Step3: Prove the inscribed - angle theorem

The measure of an inscribed angle (\(\angle DBA\)) is half the measure of the central angle that subtends the same arc (\(\overset{\frown}{DA}\)). So \(m\angle DBA=\frac{1}{2}m\overset{\frown}{DA}\).

Answer:

The proof is completed by using the properties of isosceles triangles (due to equal radii of the circle) and the exterior - angle theorem to show that the measure of the inscribed angle \(\angle DBA\) is half the measure of the arc \(\overset{\frown}{DA}\) that it subtends.