Question

given: ∠dba is inscribed in circle c, where c is outside ∠dba. prove: m∠dba = 1/2mda⌢. show hints. construct △abc and △bcd. for quadrilaterals inscribed in circles, opposite angles are supplementary, so the sum of their measures is 180°. there are two pairs of opposite angles, so the sum of the measures of the two pairs is 180°+180° = 360°. thus, by angle - sum property, inscribed angles are equal to 360°. construct two triangles △bcd and △abc, such that both are inside the circle. use the supplementary angle postulate and then addition postulate to prove the statement. construct △abc and △bcd. show that these triangles are isosceles triangles. by definition of congruent arcs, mab⌢ = max⌢, and then by definition of congruent angles, and cpctc, the inscribed angles are congruent. construct △abc and △bcd. show that these triangles are isosceles triangles. use the fact that the base angles of an isosceles triangle are the same measure, and the fact that the measure of an exterior angle equals the sum of the remote interior angles to show that the measure of the central angle is twice the measure of the inscribed angle. use the angle addition postulate to subtract the smaller angles from the larger angles.

Explanation:

Step1: Recall the properties of inscribed - angles and central - angles

The measure of a central angle is twice the measure of an inscribed angle subtended by the same arc. Let \(O\) be the center of the circle. If we consider the arc \(\overset{\frown}{DA}\), the central angle corresponding to \(\overset{\frown}{DA}\) and the inscribed angle \(\angle DBA\).

Step2: Construct isosceles triangles

Construct \(\triangle ABC\) and \(\triangle BCD\) such that \(C\) is the center of the circle. In \(\triangle ABC\) and \(\triangle BCD\), since \(CA = CB\) and \(CB=CD\) (radii of the circle), \(\triangle ABC\) and \(\triangle BCD\) are isosceles triangles. Let the central angle corresponding to arc \(\overset{\frown}{DA}\) be \(\angle DOA\) (where \(O\) is the center of the circle).

Step3: Use angle - sum properties

The sum of the measures of the angles around a point is \(360^{\circ}\). For a cyclic quadrilateral (a quadrilateral inscribed in a circle), the sum of opposite angles is \(180^{\circ}\). Let's consider the relationship between the inscribed angle \(\angle DBA\) and the central angle subtended by arc \(\overset{\frown}{DA}\).
Let the central angle of arc \(\overset{\frown}{DA}\) be \(\theta\). We know that if we take a point \(O\) as the center of the circle, and \(\angle DBA\) is an inscribed angle and \(\angle DOA\) is the central angle subtended by arc \(\overset{\frown}{DA}\).
We know that the measure of an inscribed angle \(\alpha\) and the central angle \(\beta\) subtended by the same arc are related by \(\alpha=\frac{1}{2}\beta\).
So, \(m\angle DBA=\frac{1}{2}m\overset{\frown}{DA}\)

Answer:

The proof that \(m\angle DBA = \frac{1}{2}m\overset{\frown}{DA}\) is based on the properties of inscribed angles and central angles in a circle, the construction of isosceles triangles using the radii of the circle, and the angle - sum properties of circles and cyclic quadrilaterals. The key is the relationship that the measure of an inscribed angle is half the measure of the central angle subtended by the same arc.