## Converse of Basic Proportionality Theorem

**Theorem : **

**If a point lies on a side of a triangle such that it divides the side in the ratio of the remaining sides, then that point lies on the angle bisector of the angle formed by the remaining sides.**

**Given :**

In ∆ ABC,

Point D is on side BC such that \(\large \frac {AB}{AC}\) = \(\large \frac {BD}{DC}\)

**To prove :**

Ray AD bisects ∠BAC

**Construction: **

Draw a line parallel to ray AD, passing through the point C. Extend AB so as to intersect it at point E.

**Proof :**

In ∆ BEC,

ray AD || ray CE and AC is transversal,

∴ ∠BAD ≅ ∠AEC … (i) [*corresponding angles*]

Now,

Taking AC as transversal

∠DAC ≅ ∠ACE … (ii) [*alternate angle*]

In ∆ BCE,

\(\large \frac {BA}{AE}\) = \(\large \frac {BD}{DC}\) … (iii) [*By Basic proportionality theorem*]

Also,

\(\large \frac {AB}{AC}\) = \(\large \frac {BD}{DC}\) … (iv) [*Given*]

∴ \(\large \frac {AB}{AC}\) = \(\large \frac {BA}{AE}\) …[*From equations (iii) and (iv)*]

∴ AC = AE

∴ △ACE is an isosceles triangle

∴ ∠AEC ≅ ∠ACE … (v)

∴ ∠BAD ≅ ∠DAC … [from (i), (ii) and (v)]

∴ Ray AD is the angle bisector of ∠BAC

**Hence proved**