  Selina solutions

Q12) (a) In Figure (i) BP bisects ABC and AB = AC. Find x.

(b) Find x in Figure (ii) Given: DA = DB = DC, BD bisects ABC andADB = 70°. Solution:

(a) In figure (i)

AB = AC, and BP bisects ∠ABC

AP ∥ BC is drawn

Now ∠PBC = ∠PBA

(∵ PB is the bisector of ∠ABC)

∴ AP||BC

∴ ∠APB = ∠PBC (Alternate angles)

⇒ x = ∠PBC …(i)

In ∆ ABC, ∠A = 60°

and ∠B = ∠C (∵ AB = AC)

But ∠A + ∠B + ∠C = 180°

(Angles of a triangle)

⇒ 60° + ∠B + ∠C = 180°

⇒ 60° + ∠B + ∠B = 180°

⇒ 2∠B = 180° - 60° = 120°

∠B = 120°/2 = 60°

= 1/2∠B = 60°/2 = 30° ⇒ ∠PBC = 30°

∴ From (i)

x= 30°

(b) In the figure (ii),

DA = DB = DC

BD Bisects ∠ABC

but ∠ADB + ∠DAB + ∠DBA = 180

(Angles of triangle)

⇒ 70° + ∠DBA + ∠DBA = 180° (∵ DA = DB)

⇒ 70° + 2∠DBA = 180°

⇒ 2∠DBA = 180° - 70° = 110°

∴ ∠DBA = 110°/2 = 55°

∵ BD is the bisector of ∠ABC,

∴ ∠DBA = ∠DBC, = 55°

But in ∆ DBC,

DB = DC

∴ ∠DCB = ∠DBC

⇒ X = 55°

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