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