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In DeltaA B C, D, E and F are respective...

In `DeltaA B C`, D, E and F are respectively the mid-points of sides AB, BC and CA. Show that `DeltaA B C`is divided into four congruent triangles by joining D, E and F.

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Given In a `Delta`ABC, D, E and F are respectively the mid-points of the sides AB, BC and CA.
To prove `Delta`ABC is divided into four congruent triangles.
Proof Since, ABC is a triangle and D, E and F are the mid-points of sides AB, BC and CA, respectively.
Then, `" "AD=BD=(1)/(2)AB, BE =EC=(1)/(2)BC`
and `" "AF=CF=(1)/(2)AC` ltBrgt
Now, using the mid-point theorem,
`" "EF||AB and EF=(1)/(2)AB =AD=BD`
`" "ED||AC and ED=(1)/(2)AC=AF=CF`
and `" "DF||BC and DF=(1)/(2)BC=BE=CE`
In `Delta`ADF and `Delta`EFD, `" "` AD=EF
`" "` AF=DE
and `" "` DF=FD `" "` [common]
`therefore" "DeltaADF~=DeltaEFD" "` [ by SSS congruece rule]
Similarly, `" "DeltaDEF~=DeltaEDB`
and `" "DeltaDEF~=DeltaCFE`
So, `Delta`ABC is divided into four congruent triangles. `" "` Hence proved.
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