Question
$ABC$ is a Triangle. $D$ is a point on Ab such that $\text{AD}=\frac{1}{4}\text{AD}$ and $E$ is a point on $AC$ such that $\text{AE}=\frac{1}{4}\text{AC.}$ prove that $\text{DE}=\frac{1}{4}\text{BC}.$
✓
Answer
$\triangle\text{ABC}$ is given with $D$ a point on $AB$ such that $\text{AD}=\frac{1}{4}\text{AB}.$
Also, $E$ is point on $AC$ such that $\text{AE}=\frac{1}{4}\text{AC}.$
We need to prove that $\text{DE}=\frac{1}{4}\text{BC}$ Let $P$ and $Q$ be the mid points of $AB$ and $AC$ respectively.
it is given that $\text{AD}=\frac{1}{4}\text{AB}$ and $\text{AE}=\frac{1}{4}\text{AC}$ But,
we have taken $P$ and $Q$ as the mid points of $AB$ and $AC$ respectively.
therefore, $D$ and $E$ are the mid-points of $AP$ and $AQ$ respectively.
in $\triangle\text{ABC},$ $P$ and $Q$ are the mid-points of $AB$ and $AC$ respectively.
Theorem states, the line segment joining the mid-points of any any two sides of a triangle is parallel to the third side and equal to half of it.
Therefore, we get $PQ || BC$ and $\text{PQ}=\frac{1}{2}\text{BC}\ ...(\text{i})$ in $\triangle\text{APQ},$ $D$ and $E$ are the mid-point of $AP$ and $AQ$ respectively. Therefore, we get DE || PQ and $\text{DE}=\frac{1}{2}\text{PQ}\ ...(\text{ii})$ From $(i)$ and $(ii)$,
we get: $\text{DE}=\frac{1}{4}\text{BC}$ Hence proved.
Also, $E$ is point on $AC$ such that $\text{AE}=\frac{1}{4}\text{AC}.$
We need to prove that $\text{DE}=\frac{1}{4}\text{BC}$ Let $P$ and $Q$ be the mid points of $AB$ and $AC$ respectively.
it is given that $\text{AD}=\frac{1}{4}\text{AB}$ and $\text{AE}=\frac{1}{4}\text{AC}$ But,
we have taken $P$ and $Q$ as the mid points of $AB$ and $AC$ respectively.
therefore, $D$ and $E$ are the mid-points of $AP$ and $AQ$ respectively.
in $\triangle\text{ABC},$ $P$ and $Q$ are the mid-points of $AB$ and $AC$ respectively.
Theorem states, the line segment joining the mid-points of any any two sides of a triangle is parallel to the third side and equal to half of it.
Therefore, we get $PQ || BC$ and $\text{PQ}=\frac{1}{2}\text{BC}\ ...(\text{i})$ in $\triangle\text{APQ},$ $D$ and $E$ are the mid-point of $AP$ and $AQ$ respectively. Therefore, we get DE || PQ and $\text{DE}=\frac{1}{2}\text{PQ}\ ...(\text{ii})$ From $(i)$ and $(ii)$,
we get: $\text{DE}=\frac{1}{4}\text{BC}$ Hence proved.
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