A=(7, 1)B(3, -3)are two points if the point (k=1)lines on the perpendicular bisector of AB then the value of k
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let DE cut AB extended at F
We have BE=EC
angle (BEF)= angle (DEC)
∠(FBE)=∠(ECD)
so triangle CED congruent to BEF ( case ASA)
so EF= ED and CD=BF=AB
∴ inΔ ADF, AE is the median and angle bisector of angle A so triangle ADF is isoceles.
so AD= AF=2.AB => AB= 1/2 AD
Step-by-step explanation:
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