Question
In a $\triangle\text{ABC, D}$ is the midpoint of side $AC$ such that $\text{BD}=\frac{1}{2}\text{AC}.$ Show that $\angle\text{ABC}$ is a right angle.
✓
Answer
$D$ is the mid-point of $AC$.
$\Rightarrow\text{AD = CD}=\frac{1}{2}\text{AC}$
Given, $\text{BD}=\frac{1}{2}\text{AC}$
$\Rightarrow\text{AD = CD = BD}$ Consider $\text{AD = BD}$
$\Rightarrow\angle\text{BAD}=\angle\text{ABD }...(\text{i})$ (angles opposite to equal sides are equal) Consider $\text{CD = BD}$ $\Rightarrow\angle\text{BCD}=\angle\text{CBD}...(\text{ii})$ (angles opposite to equal sides are equal) In $\triangle\text{ABC},$ angle sum property, $\angle\text{ABC}+\angle\text{BAC}+\angle\text{BCA}=180^{\circ}$
$\Rightarrow\angle\text{ABC}+\angle\text{BAD}+\angle\text{BCD}=180^{\circ}$
$\Rightarrow\angle\text{ABC}+\angle\text{ABD}+\angle\text{CBD}=180^{\circ}$ $[$From $(i)$ and $(ii)]$
$\Rightarrow\angle\text{ABC}+\angle\text{ABC}=180^{\circ}$
$\Rightarrow2\angle\text{ABC}=180^{\circ}$
$\Rightarrow\angle\text{ABC}=90^{\circ}$
Hence, $\angle\text{ABC}$ is a right angle.
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