the angle internally at the vertex. Proof Given : ABC is a triangle. AD divides BC in the ratio of the sides containing the angles Ð A to meet BC at D . That is AB AC BD DC … ( ) To prove : AD bisects Ð A i.e.
∠= ∠ Construction : Draw CE DA . Extend BA to meet at E . E O F Fig. .
(a) Fig. . (b) D D O E D Fig. .
No. Statement Reason . Let ∠ = ∠ BAD and ∠ = ∠ DAC Assumption . ∠ = ∠ = ∠ BAD AEC Since DA CE and AC is transversal, corresponding angles are equal .
∠ = ∠ = ∠ DAC ACE Since DA CE and AC is transversal, Alternate angles are equal . BA AE BD DC … ( ) In D BCE by Thales theorem . AB AC BD DC From ( ) . AB AC BA AE From ( ) and ( ) .
AC = AE … ( ) Cancelling AB . ∠= ∠ D ACE is isosceles by ( ) . AD bisects Ð A Since, ∠= ∠ = ∠= ∠ BAD DAC . Hence proved Example .
In D ABC DE BC , if , AD = , DB − , AE + and EC − then find the lengths of the sides AB and AC . Solution In D ABC we have DE BC By Thales theorem, we have AD DB AE EC x - = x gives x x Hence, x so, x = When x = , AD = , DB , AE , EC . Hence, AB AD DB , AC AE EC . Therefore, AB = , AC = .
Example . D and E are respectively the points on the sides AB and AC of a D ABC such that AB = . cm, AD = . cm, AC = .