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This is the solution.
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given
BC = AD
<ABC = <ADC
Proof
In ∆ABC & ∆ADC
1)BC= AD (GIVEN)
2)AC = AC ( COMMON SIDE)
3)<ABC = <ADC (GIVEN)
therefore ∆ABC is congruent to ∆ADC
therefore<BAC = <CAD (CPCT)
Since <BAC = <CAD, We can say that AC bisects <B
BC = AD
<ABC = <ADC
Proof
In ∆ABC & ∆ADC
1)BC= AD (GIVEN)
2)AC = AC ( COMMON SIDE)
3)<ABC = <ADC (GIVEN)
therefore ∆ABC is congruent to ∆ADC
therefore<BAC = <CAD (CPCT)
Since <BAC = <CAD, We can say that AC bisects <B
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