Question

If a line intersects sides AB and AC of a triangle ABC at D and E respectively and is parallel to BC, prove that AD/AB=AE/AC.

Answer 1

Answer:

$\frac{AD}{AB}=\frac{AE}{AC}$

Step-by-step explanation:

Concept used:

Basic proportionality theorem(Thales theorem):

If a line is drawn parallel to one side of a triangle then it cuts other two sides proportionally.

In ΔABC, DE || BC

By Thales theorem,

$\frac{AD}{DB}=\frac{AE}{EC}$

Taking reciprocals

$\frac{DB}{AD}=\frac{EC}{AE}$

Adding 1 on both sides

$1+\frac{DB}{AD}=1+\frac{EC}{AE}\\\\\frac{AD+DB}{AD}=\frac{AE+EC}{AE}\\\\\frac{AB}{AD}=\frac{AC}{AE}$

Taking reciprocals once again, we get

$\frac{AD}{AB}=\frac{AE}{AC}$

Answer 2

Answer:

Proved

Step-by-step explanation:

If a line intersects sides AB and AC of a triangle ABC at D and E respectively and is parallel to BC, prove that AD/AB=AE/AC.

DE ║ BC

in ΔABC & ΔADE

∠A = ∠A  Common Angle

∠ADE = ∠ABC   as BC ║ DE

∠AED = ∠ACB  as BC ║ DE

All three angles are equal

so

ΔABC ≅ ΔADE

in similar triangles

AD/AB =AE/AC = DE/BC

=> AD/AB =AE/AC

QED