Question

In given figure, AOB is a diameter of the circle and C, D, E are any three points on the semi-circle. Find the value of $\angle ACD+\angle BED.$

• A. ${540}^o$
• B. ${270}^o$
• C. ${180}^o$
• D. ${120}^o$

Answer 1

The correct option is B ${270}^o$

$Given-AOBEDCisapolygoninscribedinacirclewithcentreOanddiameterAOB.Tofindout-\angle ACD+\angle BED=?Solution-WejoinAE\&BC.ACDEisacyclicquadrilateral.\therefore \angle ACD+\angle AED={180}^o........\left(i\right)SimilarlyBCDEisacyclicquadrilateral.\therefore \angle BED+\angle BCD={180}^o........\left(ii\right).(\because Thesumoftheoppositepairofanglesofacyclicquadrilateral={180}^o)Again\angle ACB={90}^o=\angle AEB.........\left(iii\&iv\right)(\because Theanglesubtendedbythediameterofacircleatthecicumference={90}^o).Adding\left(i\right),\left(ii\right),\left(iii\right)\&\left(iv\right)weget\angle ACD+\angle AED+\angle BED+\angle BCD+\angle ACB+\angle AEB={180}^o+{180}^o+{90}^o+{90}^o={540}^o⟹(\angle ACD+\angle BED)+(\angle BCD+\angle BCD)+(\angle AED+\angle AEB)={540}^o⟹(\angle ACD+\angle BED)+\angle ACD+\angle BED={540}^o⟹2(\angle ACD+\angle BED)={540}^o⟹\angle ACD+\angle BED={270}^O.Ans-OptionB.$