In figure, two sides AB and BC and the median AM of a ∆ABC are respe

1.	In figure, two sides AB and BC and the median AM of a ∆ABC are respectively equal to sides DE and EF and the median DN of ∆DEF. Prove that ∆ABC ≅ ∆DEF.
Answer» ∴ AM and DN are medians of ∆ABC & ∆DEF respectively ∴ BM = MC & EN = NF ⇒ BM = 1/2 BC & EN = 1/2 EF But, BC = EF ∴BM = EN ...(i) In ∆ABM & ∆DEN we have AB = DE [Given] AM = DN [Given] BM = EN [By (i)] ∴ By SSS criterion of congruence we have ∆ABM ≅ ∆DEN ⇒ ∠B = ∠E ...(ii) [By cpctc] Now, In ∆ABC & ∆DEF AB = DE [Given] ∠B = ∠E [By (ii)] BC = EF [Given] ∴ By SAS criterion of congruence we get ∆ABC ≅ ∆DEF

In figure, two sides AB and BC and the median AM of a ∆ABC are respectively equal to sides DE and EF and the median DN of ∆DEF. Prove that ∆ABC ≅ ∆DEF.

Answer»

∴ AM and DN are medians of ∆ABC & ∆DEF respectively

∴ BM = MC & EN = NF

⇒ BM = 1/2 BC & EN = 1/2 EF

But, BC = EF ∴BM = EN ...(i)

In ∆ABM & ∆DEN we have

AB = DE [Given]

AM = DN [Given]

BM = EN [By (i)]

∴ By SSS criterion of congruence we have

∆ABM ≅ ∆DEN ⇒ ∠B = ∠E ...(ii) [By cpctc]

Now, In ∆ABC & ∆DEF

AB = DE [Given]

∠B = ∠E [By (ii)]

BC = EF [Given]

∴ By SAS criterion of congruence we get ∆ABC ≅ ∆DEF