Chapter 10.4, Problem 26PSC

To determine

To Prove : $\overline{BD}\text{bisects}\overline{\text{AC}}.$

Answer to Problem 26PSC

  $\overline{BD}\text{bisects}\overline{\text{AC}}.$

Explanation of Solution

Given information : $\overline{BD}\text{ is tangent to circle with centre A and }\overline{BD}\cong \overline{BA}.$

  

  $\begin{array}{l}\text{As AB = AD (radius) }\mathrm{.....}\text{(1)}\\ \text{and }\overline{\text{BD}}\cong \overline{AB}\text{(Given)}\\ \therefore \Delta ABD\text{ is a equilateral triangle}\text{.}\\ \text{Now }\angle \text{ABC}=90°=\angle ABD+\angle DBC\\ \angle DBC=90°-60°=30°\\ \angle BDC=180°-60°=120°\text{(Linear pair)}\\ \therefore \angle BCD=180°-120°-30°=30°\\ \Rightarrow \angle BCD=\angle DBC(30°)\\ \therefore BD=DC\mathrm{....}(2)\text{(Sides are equal to opposite equal angles)}\\ \therefore \text{By}\text{(1)}\text{and (2)}\\ \text{AD}=DC\Rightarrow BD\text{ bisect AC}\end{array}$