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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