Question

what is the measure of angle abd? what circle property rule did you use? a d 17 cm c b 15 cm

Explanation:

Step1: Find the length of \(BC\) and \(CD\)

Since \(BC\) and \(CD\) are radii of the circle, \(BC = CD=15\mathrm{cm}\), and \(AD = 17\mathrm{cm}\), so \(AC=CD = 15\mathrm{cm}\), \(AD\) is the diameter of the circle.

Step2: Apply the Pythagorean theorem in \(\triangle ABD\)

By the property that an angle inscribed in a semicircle is a right - angle (\(\angle ABD\) is inscribed in a semicircle with diameter \(AD\)). Let's confirm using the Pythagorean theorem. First, find \(AB\) using the Pythagorean theorem in \(\triangle ABD\). We know that \(AD = 17\mathrm{cm}\), \(BD = 15\mathrm{cm}\). Let \(AB=x\). Then \(x^{2}+BD^{2}=AD^{2}\) (by Pythagoras, since \(\angle ABD = 90^{\circ}\) if the circle property is used directly, but for verification):

$$x^{2}+15^{2}=17^{2}$$

$$x^{2}=17^{2}-15^{2}=(17 + 15)(17 - 15)=32\times2=64$$

$$x = 8\mathrm{cm}$$

Answer:

The measure of \(\angle ABD\) is \(90^{\circ}\). The circle property rule used is: An angle inscribed in a semicircle is a right - angle.