Question

In triangle ABC if AD is the bisector of angleA prove that Area of triangle ABD /area of triangle ACD = AB/AC

Answer 1

$\frac{area \bigtriangleup ABD}{area \bigtriangleup ACD} = \frac{AB}{AC}$

Step-by-step explanation:

in triangle ABC ,  AD is the bisector of angleA

let us draw a perpendicular

AL ⊥ BC

then area of triangle ABD = $\frac{1}{2}\times base \times height$

= $\frac{1}{2} \times BD \times AL$

area of triangle ACD = $\frac{1}{2} \times CD \times AL$

then their ratio is

$\frac{area \bigtriangleup ABD}{area \bigtriangleup ACD} = \frac{\frac{1}{2} \times BD \times AL}{\frac{1}{2} \times CD\times AL}$

$\frac{area \bigtriangleup ABD}{area \bigtriangleup ACD} = \frac{BD}{CD}$..(1)

since , AD is the bisector of angle A

then ,

$\frac{AB}{AC} = \frac{BD}{CD}$..(2)

From 1 and 2

$\frac{area \bigtriangleup ABD}{area \bigtriangleup ACD} = \frac{AB}{AC}$

HENCE PROVED

#Learn more:

ABC is an isosceles triangle in which Ab =AC and Ad is the bisector of angle a. is triangle Abd  congruent to triangle acd

https://brainly.in/question/7829526

Answer 2

In △ABC, AD is the bisector of ∠A.

∴ AB/AC = BD/DC ...(i)

From A, draw AL ⊥ BC.

∴ Area(△ABD)/Area(△ACD) = (1/2)BD.AL/(1/2)DC.AL

⇒ Area(△ABD)/Area(△ACD) = BD/DC

⇒ Area(△ABD)/Area(△ACD) = AB/AC ...[From(i)]     [Hence proved]