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D, E and F are respectively the mid-points of the sides AB, BC and CA of a triangle ABC. Prove that by joining these mid-points D, E and F, the triangles ABC is divided into four congruent triangles.

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Solution.
 

Given: In \triangle ABC, D, E and F are respectively the mid-points of the sides AB, BC and CA.
To prove: \triangle ABC is divided into four congruent triangles.

Proof: Using given conditions we have
AD=BD=\frac{1}{2}AB,BE=EC=\frac{1}{2}BC,AF=CF=\frac{1}{2}AC
Using mid-point theorem
EF\parallel AB  and   EF=\frac{1}{2}AB=AD=BD
ED\parallel AC  and  ED=\frac{1}{2}AC=AF=CF
DF\parallel BC  and   DF=\frac{1}{2}BC=BE=EC

In \triangle ADF and \triangle EFD
 AD = EF
 AF = DE
 DF = FD {common side}
\triangle ADF\cong \triangle EFD  
{by SSS congruence}
Similarly we can prove that,
\triangle DEF\cong \triangle EDB
\triangle DEF\cong \triangle CFE

So, \triangle ABC is divided into four congruent triangles
Hence proved

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