# ABC is a triangle.D is a point on AB,such that AD=1/4 AB and E is a point on AC such that AE =1/4 AC . Prove that : DE=1/4 BC

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You need to prove that the two traingles are similar triangles.

`AD= 1/4AB` (given)

`AE = 1/4 AC` (given)

A is a common angle in `DeltaABC and DeltaADE`

`therefore (AD)/(AB)=(AE)/(AC)`

`therefore (DE)/(BC)=(AD)/(AB)=(AE)/(AC)`

`therefore Delta ABC similar to Delta ADE` ( SAS)

As the ratio of the other 2 sides is 1:4 and these are similar triangles

`therefore` **DE=1/4 BC**

It is given that ABC is a triangle with point D on AB such that AD=(1/4)*AB and point E on AC such that AE = (1/4)*AC .

Two triangles are similar if the length of two corresponding sides is in the same proportion and the included angle between the sides is the same. In triangles ABC and ADE the length of sides AD and AB and sides AE and AC is in the same proportion and angle A is common for both of them; this makes the two triangles similar. In similar triangles the ratio of the lengths of corresponding sides is the same; as a result AD/AB = AE/AC = DE/BC = 1/4.

**This gives DE = (1/4)*BC**

Thank you teachers! for your efforts and now the question is clear! thank you!!