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Answer:
Step-by-step explanation:
Given: △ABC, BC>AC, D∈ AC , CD=CB
To prove: m∠ABD is acute
Proof: In ΔABC, the angle opposite to side BC is ∠BAC and the angle opposite to side AC is ∠ABC.
Now, it is given that BC>AC, then ∠BAC>∠ABC.. (1)
In ΔBDC, using the exterior angle property,
∠ADB=∠DBC+∠BCD
∠ADB=∠DBC+∠BCA
⇒∠ADB>∠BAC (2)
From equation (1) and (2), we get
∠ADB>∠BAC
⇒∠ADB>∠ABC
⇒DB>AB
Hence, m∠ABD is acute
