in a triangle ABC , AD bisects LA. prove that AC >CD
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Answered by
2
Answer:
What is LA here? Make it clear.
Answered by
2
Answer:
Step-by-step explanation:
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.
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