Fig. 11.38
13. In a A ABC, AD bisects ZA and ZC > ZB. Prove that ZADB > ZADC.
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Step-by-step explanation:
"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."
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