Question

In triangle abc angle b is twice of angle c and bisector of angle b intersects ac at d prove that bd/ad=bc/ab

Answer 1

$\dfrac{BD}{AD}=\dfrac{BC}{AB}$

Hence proved

Step-by-step explanation:

Given: In ΔABC, ∠B is twice of ∠C. The bisector of ∠B intersects AC at D.

To prove: $\dfrac{BD}{AD}=\dfrac{BC}{AB}$

Figure: In attachment

In ΔBDC

∠DBC = ∠DCB         (Given)

BD = DC  ( Opposite sides of equal angle is equal)

In ΔABC, BD is angle bisector of ∠ABC

$\dfrac{CD}{AD}=\dfrac{BC}{AB}$

Angle bisector theorem of triangle: The bisector of vertex angle divide base into two part then the ratio of sides to base division is equal.

But  CD = BD  

Therefore,

$\dfrac{BD}{AD}=\dfrac{BC}{AB}$

Hence proved

#Learn more:

https://brainly.in/question/8571549