Question

In the given figure, BD = AD and DC = AC, If $\angle ABC={35}^{\circ },$ then $\angle ACD=?$

• A. ${30}^{\circ }$
• B. ${35}^{\circ }$
• C. ${40}^{\circ }$
• D. ${45}^{\circ }$

Answer 1

The correct option is C

${40}^{\circ }$

$\underset{–}{In\Delta ABD}$

(i) AD = BD ...(given)

(ii) $\therefore \angle DBA=\angle BAD={35}^{\circ }...$ (base angles of an isosceles $\Delta$)

(iii) $\therefore \angle ADC={70}^{\circ }...$ (exterior angle theorem)

$\underset{–}{In\Delta ADC}$

(iv) DC = AC...(given)

(v) $\therefore \angle CDA=\angle DAC={70}^{\circ }...$ (base angles of an isosceles $\Delta$)

(vi) $\therefore \angle ACD={180}^{\circ }-({70}^{\circ }+{70}^{\circ })...$ (Sum of LS of a $\Delta$)

$={180}^{\circ }-{140}^{\circ }$

= ${40}^{\circ }$