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Based on the diagram above, the following relation holds true.
BC>B′C′⇒m∠BAC>m∠B′A′C′
Consider △ABC and △A′B′C such that AB=A′B′ and AC=A′C′, where BC>B′C′.
This theorem can be proven by contradiction. Since the goal is to prove that m∠BAC>m∠B′A′C′ the opposite statement will be assumed, that is, m∠BAC≤m∠B′A′C′. Because this is a non-strict inequality, both m∠BAC=m∠B′A′C′ and m∠BAC<m∠B′A′C′ have to be considered.
If m∠BAC<m∠B′A′C′ then the Hinge Theorem states that BC<B′C′, but this also contradicts the fact that BC>B′C′.
The assumption that m∠BAC is less than or equal to m∠B′A′C′ contradicts the hypothesis. Therefore, this assumption must be false. Consequently, the initial conclusion of the theorem is true.
m∠BAC>m∠B′A′C′.