Question

Please solve this from Chapter - Similarities
Plzz Dont use bisector theorem here

Answer 1

Answer:

Step-by-step explanation:

given: ABC is a triangle.

∠ABC = 2∠ACB ..........(1)

and BP is angle bisector of ∠ABC.

TPT : CB : BA = CP : PA

proof:

let ∠ ACB = x

therefore ∠ABC = 2x

since BP is angle bisector of ∠ABC.

∠ABP = ∠PBC = x

∠ BPA = ∠PBC +∠BCP [exterior angle is equal to the sum of  opposite interior angles]

∠BPA = x+x =2x.............(2)

now in the triangle ABC and triangle ABP,

∠ACB = ∠ABP

∠ABC = ∠APB

therefore ΔABC ≈  ΔABP

thus by the similar triangle property ratio of corresponding sides are equal.

hence

CB/BA=BP/PA....(3)

BP = CP [sides opposite to equal angles are equal]

thus

CB/BA=CP/PA

⇒CB : BA = CP : PA

Answer 2

Answer:
Step-by-step explanation:
given: ABC is a triangle.
∠ABC = 2∠ACB ..........(1)
and BP is angle bisector of ∠ABC.
TPT : CB : BA = CP : PA
proof:
let ∠ ACB = x
therefore ∠ABC = 2x
since BP is angle bisector of ∠ABC.
∠ABP = ∠PBC = x
∠ BPA = ∠PBC +∠BCP [exterior angle is equal to the sum of  opposite interior angles]
∠BPA = x+x =2x.............(2)
now in the triangle ABC and triangle ABP,
∠ACB = ∠ABP
∠ABC = ∠APB
therefore ΔABC ≈  ΔABP
thus by the similar triangle property ratio of corresponding sides are equal.
hence
CB/BA=BP/PA....(3)
BP = CP [sides opposite to equal angles are equal]
thus
CB/BA=CP/PA
⇒CB : BA = CP : PA

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