Q1 of 56 Page 6

The bisector of the exterior angle ∠A of Δ ABC intersects side BC produced at D. Prove that .
                     

Given: ABC is a triangle; AD is the exterior bisector of ∠A and meets BC produced at D; BA is produced to F.
To prove:                 
Construction: Draw CE||DA to meet AB at E.
Proof: In Δ ABC, CE || AD cut by AC.
  ∠CAD = ∠ACE (Alternate angles)
Similarly CE || AD cut by AB
  ∠FAD = ∠AEC (corresponding angles)
Since ∠FAD = ∠CAD (given)
     ∴ ∠ACE = ∠AEC
\\ AC = AE ( by isosceles Δ theorem)
Now in Δ BAD, CE || DA
AE = DC (BPT)
AB    BD
But AC = AE (proved above) 
\\ AC = DC
 AB     BD    or (proved).

More from this chapter

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2 State and prove the converse of angle bisector theorem.
                                3 If a parallelogram has all its sides equal and one of its diagonal is equal to a side, show that its diagonals are in the ratio : 1. 4 In Δ ABC, P, Q are points on AB and AC respectively and PQ || BC. Prove that the median AD bisects PQ.
                              5 Prove that the line joining the midpoints of the sides of the triangle form four triangles, each of which is similar to the original triangle.