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Question
In a rectangle ABCD, AB = 25 cm and BC = 15. In what ratio does the bisector of ∠C divide AB?
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Solution
Given, AB = 25 cm and BC = 15 cm
Now, In rectangle ABCD,
CO is the bisector of ∠C and it divides AB.
∴ ∠OCB = ∠OCD = 45°
∴ ∠OCB = ∠OCD = 45°
In ΔOCB, we have
∠CBO + ∠OCB + ∠COB = 180° ...[Angle sum property of triangle]
90° + 45° + ∠COB = 180°
∠COB = 180° – 90° + 45°
∠COB = 180° – 135° = 45°
Now, In ΔOCB,
∠OCB = ∠COB
Then, OB = OC
⇒ OB = 15 cm
CO divides AB in the ratio AO : OB
Let AO be x, then OB = AB – x = 25 – x
Hence, AO : OB = x : 25 – x
⇒ 10 : 15
⇒ 2 : 3
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