In the adjoining figure

(i) AB=BC, M is the mid -point of AB and N is the mid- point of BC. Show that AM=NC.
(ii) BM=BN,M is the mid- point of AB and N is the mid - point of BC. Show that AB=BC.

Explanation

Given, AB=BC …(i)
M is the mid - point of AB.
`AM=MB=(1)/(2)AB` …(ii)
and N is the mid - point of BC.
`BN=NC=(1)/(2)BC` …(iii)
According to Euclid's axiom , things which are halves of the same things are equal to one another . From Eq. (i) , AB=BC
On multiplying both sides by `(1)/(2)` , we get
`(1)/(2)AB=(1)/(2)BC`
`rArr AM=NC` `["using Eqs. (ii) and (iii)"]`
(ii) Given, BM=BN ...(i)
M is the mid -point of AB.
`:. AM=BM=(1)/(2)AB`
`rArr2AM=2BM=AB` ...(ii)
and N is the mid - point of BC.
`:. BN=NC=(1)/(2)BC`
`rArr 2BN=2NC=BC` ...(iii)
According to Euclid's which are double of the same thing are equal to one another.
On multiplying both sides of Eq.(i) by 2, we get
2BM=2BN
`rArr AB=BC ` ` ["using Eqs. (ii) and (iii)"]`