1 : 1 Solution: In $\triangle\text{ABC}$ AB = AC $\therefore\ \angle\text{ABC} = \angle\text{ACB}$ (angles opposite to equal sides of a triangle are equal) ...(1) In $\triangle\text{DBC}$ DB = DC, $\therefore\ \angle\text{DBC} = \angle\text{DCB}$ (angles opposite to equal sides of a triangle are equal) ...(2) Subtract 2 from 1 $\angle\text{ABC} - \angle\text{DBC} = \angle\text{ACB} - \angle\text{DCB}$ (equals subtracted from equals gives equal) $= \angle\text{ABD} = \angle\text{ACD}$ Divide both the sides by $\angle\text{ACD}$ $\Rightarrow \frac{\angle\text{ABD}}{\angle\text{ACD}} = 1$ $\therefore\ \angle\text{ABD} : \angle\text{ACD}=1:1$
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