Question

(0)
ZABC and ZADB = 70°.
Р
60
70
B4
С
BEC is an equilateral triangle
A
B
12. (a) In Figure (0) BP bisects ZABC and AB = AC. Find x.
(0) Find x in Figure (ii).
Given : DA = DB = DC. BD bisects
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Answer 1

Answer:

From the figure

AB = AC and BP bisects ∠ABC

AP is drawn parallel to BC

Here PB is the bisector of ∠ABC

∠PBC=∠PBA

∠APB=∠PBC are alternate angles

x=∠PBC (1)

In Δ ABC

∠A=60

Since AB = AC we get ∠B=∠C

In a triangle

∠A+∠B+∠C=180

Substituting the values

60∠+∠B+∠C=180∠

We get

60∠+∠B+∠B=180∠

By further calculation

2∠B=180−60=120

∠B=120/2=60

∠B=60/2=30

∠PBC=30

M is any point on the side BC of ΔABC in such a way that AM is the bisector of ∠BAC. Is it true to say that perimeter of ΔABC is greater than 2AM?

Answer 2

Answer:

the above answer is correct

you can copy that

I hope it help you

Step-by-step explanation:

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