In the figure,O is the centre of a circle and BCD is tangent to it at C. Prove that ∠BAC + ∠ACD = 90°.

Question

I am a beginner to these type of questions relating to tangents ..... Thats all.

Answer 1

$BC$ is tangent, we have $$\angle OCD=90^\circ$$

$OA=OC$, so $\angle OAC=\angle OCA$. Hence

$$\angle BAC+\angle ACD= \angle OCA+\angle ACD=\angle OCD=90^\circ$$

Answer 2

HINT : By alternate segment theorem, $$\angle PCB=\angle BAC.$$ Also, since two triangles $COA, COP$ are isosceles triangles, we have $$\angle OCA=\angle OAC, \angle POC=2\angle PCB.$$

Hence, $$\begin{align}\angle{BAC}+\angle{ACD}&=\angle OCA+\angle ACD\\&=180^\circ-(\angle PCB+\angle PCO)\\&=180^\circ -\{\angle PCB+(180^\circ-2\angle PCB)/2\}\\&=180^\circ-90^\circ\\&=90^\circ.\end{align}$$