In the Fig.5.12:
(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.
i. Given, AB = BC …(i)
M is the mid-point of AB
∴ AM = MB = 
…(ii)
and N is the mid-point of BC.
∴ BN = NC = 
…(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 
,
We get,
⇒ AM = NC [using Eq. (ii) and (iii)]
ii. Given, BM =BN …(i)
M is the mid-point of AB.
∴ AM = BM = 
AB
⇒ 2AM = 2BM = AB …(ii)
and N is the mid-point of BC.
∴ BN = NC = 
BC
⇒ 2BN = 2NC = BC …. (iii)
According to Euclid’s axiom, things 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
⇒ AB = BC [using Eq. (i) and (ii)]