In the Fig,

(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)]