Draw the internal bisector on each angle for a triangle of each side
Answers
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)
solution
I hope you are understand my solution
Answer:
Ram ram mithe
Step-by-step explanation:
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)
solution