Question

In triangle abc, ad bisects angle a and angle c greater than angle


b. prove that angle adb greater than angke abc

Answer 1

Answer:

The proof is explained step-wise below :

Step-by-step explanation:

Given : m∠C > m∠B and AD bisects ∠A

To Prove : m∠ADB > m∠ABC

Proof : Since m∠C > m∠B

⇒ m∠ACB > m∠ABC

⇒ m∠ACB + m∠CAD > m∠ABC + m∠BAD .....(1)

(AD bisects ∠A ⇒m∠CAD = m∠BAD )

In ΔABD,

m∠ABC + m∠BAD + m∠ADB = 180°

Therefore, m∠ABC + m∠BAD = 180° – ∠ADB ....(2)

In ΔACD,

m∠ACD + m∠CAD + m∠ADC = 180°

Therefore, m∠ACB + m∠CAD = 180° – m∠ADC ....(3)

From (1), (2) and (3), we get

180° – m∠ADC > 180° – m∠ADB

⇒ m∠ADB – m∠ADC > 180° – 180°

⇒ m∠ADB – m∠ADC > 0

⇒ m∠ADB > m∠ADC

Hence proved.

Answer 2

"In ΔABC,

∠C > ∠B (Given)

∴ ∠ACB > ∠ABC

⇒ ∠ACB + ∠2 > ∠ABC + ∠1 .....(1) (AD bisects ∠A ⇒∠1 = ∠2)

In ΔABD,

∠ABC + ∠1 + ∠ADB = 180°

∴ ∠ABC + ∠1 = 180° – ∠ADB ....(2)

In ΔACD,

∠ACD + ∠2 + ∠ADC = 180°

∴ ∠ACB + ∠2 = 180° – ∠ADC ....(3)

From (1), (2) and (3), we get

180° – ∠ADC > 180° – ∠ADB

∴ ∠ADB – ∠ADC > 180° – 180°

⇒ ∠ADB – ∠ADC > 0

⇒ ∠ADB > ∠ADC

Hence proved."