Question

P is an interior point of Z ABC. If Z ABC = 80º,
PBC = 46° and CB is produced to D, find the
measure of the angle between BP and the bisector
of the exterior angle of Z ABC.

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Answer 1

Answer:

ANSWER

∠ABP=

2

1

∠B (BP bisects ∠B)

∠BAP=∠A+

2

1

(Ext.∠A) (AP bisects ∠A)

∠BAP=∠A+

2

1

(180−∠A)

∠BAP=90+

2

1

∠A

In △ABP,

∠BAP+∠PBA+∠APB=180 (Sum of angles of triangle)

90+

2

1

∠A+

2

1

∠B+∠APB=180

∠APB=90−

2

1

(∠A+∠B)

2∠APB=180−∠A−∠B

2∠APB=∠C (Angles sum property on triangle ABC)

Step-by-step explanation:

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