O is the centreof the circle and BCD is a tangent to it at C. Prove that angle ABC + angle ACD= 90°

O is the centreof the circle and BCD is a tangent to it at C. Prove th

1.	O is the centreof the circle and BCD is a tangent to it at C. Prove that angle ABC + angle ACD= 90°
Answer» {tex}\\angle OCD = 90^\\circ{/tex} (tangent and radii are {tex}\\bot {/tex} to one another at the point of contact)In {tex}\\triangle{/tex}OCA,OC = OA (radii of circle)Hence, {tex}\\angle OCA = \\angle OAC{/tex} (angles opposite to equal sides are equal)Also, {tex}\\angle OCD = \\angle OCA + \\angle ACD{/tex}{tex}90^\\circ = \\angle OAC + \\angle ACD{/tex}{tex}\\left( {\\because \\angle OCA = \\angle OAC} \\right){/tex}{tex}90^\\circ = \\angle BAC + \\angle ACD{/tex}Hence, {tex}\\angle BAC + \\angle ACD = 90^\\circ{/tex}Hence proved.