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In a δ Abc, Ad Bisects ∠A and ∠C > ∠B. Prove that ∠Adb > ∠Adc. - Mathematics

Answer in Brief

In a Δ ABC, AD bisects ∠A and ∠C > ∠B. Prove that ∠ADB > ∠ADC.

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Solution

In the given ΔABCAD bisects ∠Aand ∠C >∠B. We need to prove ∠ADB >∠ADC.

 

Let,

 ∠BAD = ∠1

∠DAC = ∠2

∠ADB = ∠3

∠ADC = ∠4

Also, 

As AD bisects ∠A

∠1 = ∠2…..(1) 

Now, in ΔABD, using exterior angle theorem, we get,

∠4 = ∠B + ∠1

Similarly,

∠3 = ∠2 + ∠C

∠3 = ∠1 + ∠C  [using (1)]

Further, it is given,

∠C >∠B

Adding ∠1to both the sides

∠C +∠1 >∠B + ∠1

∠3 > ∠4

Thus, ∠3 > ∠4

Hence proved.

  Is there an error in this question or solution?
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APPEARS IN

RD Sharma Mathematics for Class 9
Chapter 11 Triangle and its Angles
Exercise 11.2 | Q 13 | Page 22
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