Keerthana
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मेंABC,D\triangle में A B C, D और EE क्रमशः ABA B और AC\mathrm{AC} पक्षों के दो मध्य बिंदु हैं। अगर BAC=40\angle \mathrm{BAC}=40^{\circ} और ABC=\angle \mathrm{ABC}=

6565^{\circ} तो CED\angle \mathrm{CED} है:

a

130°

b

75°

c

125°

d

105°

Answer : Option D
Explanation :

BAC=40\angle \mathrm{BAC}=40^{\circ}

ABC=65\angle \mathrm{ABC}=65^{\circ}

ACB=1804065=75\therefore \angle \mathrm{ACB}=180^{\circ}-40^{\circ}-65^{\circ}=75^{\circ}

DEBC\mathrm{DE} \| \mathrm{BC}

AED=ACB=75\therefore \quad \angle \mathrm{AED}=\angle \mathrm{ACB}=75^{\circ}

CED=18075=105\therefore \quad \angle \mathrm{CED}=180^{\circ}-75^{\circ}=105^{\circ}

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