In a rectangle ABCD , AB = 25 cm and BC = 15 cm . in what ratio does the bisector of angle C divide AB?

AB = 25 cm, BC = 15 cm
CE is the bisector of ∠C = ∠BCE = 45°
Therefore, ∠BEC = 45° {Angle sum property of a triangle]
 BC = BE  [Sides opposite to equal angles are equal]

 BE = 15 cm

AE = 25 - 15 = 10 cm

AE:BE = 10 : 15 = 2 : 3

Therefore, the ratio in which the bisector of ∠C divides AB is 2:3.
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Sol:

AB = 25 cm, BC = 15 cm
CE is the bisector of ∠C ⇒ ∠BCE = 45°
⇒ ∠BEC = 45° {Angle sum property of a triangle]
⇒ BC = BE [Sides opposite to equal angles are equal]

⇒ BE = 15 cm

AE = 25 - 15 = 10 cm

AE:BE = 10 : 15 = 2 : 3

Therefore, the ratio in which the bisector of ∠C divides AB is 2:3.
  • 5
AB = 25 cm, BC = 15 cm
CE is the bisector of ∠C = ∠BCE = 45°
Therefore, ∠BEC = 45° {Angle sum property of a triangle]
 BC = BE  [Sides opposite to equal angles are equal]

 BE = 15 cm

AE = 25 - 15 = 10 cm

AE:BE = 10 : 15 = 2 : 3

Therefore, the ratio in which the bisector of ∠C divides AB is 2:3.
  • 0
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