Question

In the adjoining figure, $AB=AC=CD$, $\angle ADC={35}^o$. Then (i) $\angle ABC$, (ii) $\angle BEC$ are respectively:

• A. ${110}^o\&{70}^o.$
• B. ${350}^o\&{1100}^o.$
• C. ${50}^o\&{35}^o.$
• D. ${70}^o\&{40}^o.$

Answer 1

Answer: (D)

${70}^o\&{40}^o.$

$Given-ABCEisacyclicquadrilateral.AE\&BCareproducedtomeetatD.AC\&EBarejoined.AB=AC=CDand\angle ADC={35}^o.Tofindout-\left(i\right)\angle ABC=?\left(ii\right)\angle BEC=?Solution-In\Delta ACD,AC=CD⟹\Delta ACDisisosceleswithADasbase.So\angle CAD=\angle CDA={35}^o.i.e\angle CAD+\angle CDA=2\angle CAD=2\times {35}^o={70}^o.\therefore \angle ACD={180}^o-2\angle CAD={180}^o-{70}^o={110}^o.\left(anglesumpropetyoftriangles\right).\therefore \angle ACB={180}^o-\angle ACD={180}^o-{110}^o={70}^o.\left(linearpair\right)NowAC=AB⟹\Delta ABCisisosceleswithBCasbase.So\angle ABC=\angle ACB={70}^o.i.e\angle ABC+\angle ACB=2\angle ABC=2\times {70}^o={140}^o.\therefore \angle BAC={180}^o-2\angle ABC={180}^o-{140}^o={40}^o.\left(anglesumpropetyoftriangles\right).Now\angle BAC\&\angle BEChavebeensubtendedbythechordBCtothecircumferenceofthecircleatA\&E.\therefore \angle BAC=\angle BEC={40}^o.So,\angle ABCand\angle BECarerespectively,{70}^o\&{40}^o.Hence,optionDiscorrect.$