Question

5. In A ABC, angle ABC is equal to twice the
angle ACB, and bisector of angle ABC meets
the opposite side at point P. Show that:
(i) CB : BA = CP : PA
(ii) AB X BC = BP X CA​

Answer 1

Answer:

In ΔABC, ∠ABC = 2∠ACB

Let ∠ACB = x

⇒∠ABC = 2∠ACB = 2x

Given BP is bisector of ∠ABC

Hence ∠ABP = ∠PBC = x

By angle bisector theorem the bisector of an angle divides the side opposite to it in the ratio of other two sides.

Hence AB:BC = CP:PA

2) Consider ΔABC and ΔAPB

∠ABC = ∠APB [Exterior angle property]

∠BCP = ∠ABP [Given]

∴ ΔABC ≈ ΔAPB [AA criterion]

∴ AB / BP = CA / CB [Corresponding sides of similar triangles are proportional.]

⇒ AB x BC = BP x CA.

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Step-by-step explanation:

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Answer 2

Given:

• ABC is a Triangle
• 2∠ACB=∠ABC
• Bisector of ∠ABC meets AC at point P.

To prove:

1. CB : BA = CP : PA
2. AB X BC = BP X CA

Proof:

1) In ∆ABC,

∠ABC=2∠ACB

Let ∠ABC be x

So, ∠ABC=2x

And, BP is bisector of ∠ABC

⇛ ∠ABP=∠PCB=x

$\boxed{</p><p></p><p>\begin{minipage}{5 cm}</p><p></p><p>Angle \:bisector \:theorem- \\The \:angle bisector \:of \:an\: angle \\ divides \:the \:side\: opposite\: to \\ it\: in \:the \:ratio \:of \:other\: two\\ sides.\theta </p><p></p><p>\end{minipage}</p><p></p><p>}$

Hence, AB:BC=CP:PA

2) in ∆ABC and ∆APB

∠BCP=∠ABP $\underbrace{exterior \: angle}$

∠BCP=∠ABP $\underbrace{Given}$

∴∆ABC≅∆APB $\underbrace{AA\:Criteria}$

Corresponding part of congurent triangle [CPCT]

$\frac{AB}{BP}$=$\frac{BP}{CA}$

∴AB×BC=BP×CA

$\bold\green{Thankyou!}$