prove that the internal bisector of an angle of a traingle divides the opposite side internally in the ratio of the other two sides.4 mark answer please it's urgent
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The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the corresponding sides containing the angle. Prove it.
Step-by-step explanation:
ANSWER
Given:
Let ABC be the triangle
AD be internal bisector of ∠BAC which meet BC at D
To prove:
DC
BD
=
AC
AB
Draw CE∥DA to meet BA produced at E
Since CE∥DA and AC is the transversal.
∠DAC=∠ACE (alternate angle ) .... (1)
∠BAD=∠AEC (corresponding angle) .... (2)
Since AD is the angle bisector of ∠A
∴∠BAD=∠DAC .... (3)
From (1), (2) and (3), we have
∠ACE=∠AEC
In △ACE,
⇒AE=AC
(∴ Sides opposite to equal angles are equal)
In △BCE,
⇒CE∥DA
⇒
DC
BD
=
AE
BA
....(Thales Theorem)
⇒
DC
BD
=
AC
AB
....(∴AE=AC)
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