Question

ABC is a triangle, D is a point on AB such that AD = $\frac{1}{4}$ AB and E is a point on AC such that AE = $\frac{1}{4}$ AC. Prove that DE = $\frac{1}{4}$ BC.

Answer 1

is given with D a point on AB such that .

Also, E is point on AC such that.

We need to prove that

Let P and Q be the mid points of AB and AC respectively.

It is given that

and

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 , P and Q are the mid-points of AB and AC respectively.

Theorem states, the line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it.

Therefore, we get and …… (i)

In , D and E are the mid-points of AP and AQ respectively.

Therefore, we get and …… (ii)

From (i) and (ii),we get:

Hence proved.