Statement: Line drawn parallel to the any side of a triangle cuts the remaining two sides in the same ratio.

Given:

Line XY is drawn parallel to the base BC of a triangle ABC and meets the points E and F of sides AB and AC respectively.

To prove:

AE
EB
=
AF
FC

Setup: (Click me)

Identify triangles ABC and AEF.

Proof: (Click me)

S. No. Fact S. No. Reason
1. In ΔABC and ΔAEF 1. Separation for comparison
(i) ∠ABC = ∠AEF (A) (i) Corresponding angles (XY || BC)
(ii) ∠ACB = ∠AFE (A) (ii) Corresponding angles (XY || BC)
(iii) ∠BAC = ∠EAF (A) (iii) Common angle
2. ΔABC ~ ΔAEF 2. All corresponding angles are equal
3.
AB
AE
=
AC
AF
AE + EB
AE
=
AF + FC
AF
1 +
EB
AE
= 1 +
FC
AF
 3. Ratio of corresponding sides of similar triangles
4.
EB
AE
=
FC
AF
AE
EB
=
AF
FC
 4. By simplifying and taking reciprocal
Proved.