Question

In the figure $D$and$E$ are the mid-points of sides $AB$and$AC$ respectively of $\Delta ABC$. Find $\angle EDB.$

• A. ${110}^{\circ }$
• B. ${120}^{\circ }$
• C. ${70}^{\circ }$
• D. ${80}^{\circ }$

Answer 1

Answer: (A)

${110}^{\circ }$

In $△ABC$

$\Rightarrow$ $\angle A+\angle B+\angle C={180}^o$

$\Rightarrow$ ${60}^o+\angle B+{50}^o={180}^o$

$\Rightarrow$ ${110}^o+\angle B={180}^o$

$\Rightarrow$ $\angle B={70}^o$

$\therefore$ $\angle ABC={70}^o$ ----- ( 1 )

In $△ADE$ and $△ABC$,

$\Rightarrow$ $AB=2AD$ [ $D$ is mid-point of $AB$ ]

$\therefore$ $\frac{AD}{AB}=\frac{1}{2}$ ---- ( 2 )

$\Rightarrow$ $AC=2AE$ [ $E$ is a mid-point of side $AC$ ]

$\therefore$ $\frac{AE}{AC}=\frac{1}{2}$ ---- ( 3 )

$\Rightarrow$ $\frac{AD}{AB}=\frac{AE}{AC}=\frac{1}{2}$ [ From ( 1 ) and ( 2 ) ]

$\Rightarrow$ $\angle A=\angle A$ [ Common angle ]

$\therefore$ $△ADE\sim △ABC$ [ By $SAS$ similarity rule ]

$\Rightarrow$ $\angle ADE=\angle ABC$ [ Corresponding angles of similar triangle are equal ]

$\therefore$ $\angle ADE={70}^o$ [ From ( 1 ) ]

$\Rightarrow$ $\angle ADE+\angle EDB={180}^o$ [ Linear pair ]

$\Rightarrow$ ${70}^o+\angle EDB={180}^o$

$\Rightarrow$ $\angle EDB={110}^o$.