show that the internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle
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☆ Given : A ΔABC in which AD is the internal bisector of ∠A and meets BC in D.
Prove that : BD / DC = AB / AC
Construction : Draw CE || DA to meet BA produced in E.
StatementsReasons1) CE || DA
1) By construction2) ∠2 = ∠32) Alternate interior angles3) ∠1 = ∠43) Corresponding angles4) AD is the bisector4) Given5) ∠1 =∠25) Definition of angle bisector6) ∠3= ∠46) From (2) and (3)7) AE = AC7) In ΔACE, side opposite to equal angles are equal8) BD / DC = BA / AE8) In ΔBCE DA || CE and by BPT theorem9) BD / DC = AB / AC9) From
☆ ☆ ☆ Hop its helpful ☆ ☆ ☆
☆ Regards :- ♡♡《 Nitish kr singh 》♡♡
Solution :-
☆ Given : A ΔABC in which AD is the internal bisector of ∠A and meets BC in D.
Prove that : BD / DC = AB / AC
Construction : Draw CE || DA to meet BA produced in E.
StatementsReasons1) CE || DA
1) By construction2) ∠2 = ∠32) Alternate interior angles3) ∠1 = ∠43) Corresponding angles4) AD is the bisector4) Given5) ∠1 =∠25) Definition of angle bisector6) ∠3= ∠46) From (2) and (3)7) AE = AC7) In ΔACE, side opposite to equal angles are equal8) BD / DC = BA / AE8) In ΔBCE DA || CE and by BPT theorem9) BD / DC = AB / AC9) From
☆ ☆ ☆ Hop its helpful ☆ ☆ ☆
☆ Regards :- ♡♡《 Nitish kr singh 》♡♡
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