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Question
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 a ΔABC, D, E and F are respectively the mid-points of the sides AB, BC and CA.
To prove: ΔABC 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
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 ΔADF and ΔEFD,
AD = EF
AF = DE
And DF = FD ...[Common]
∴ ΔADF ≅ ΔEFD ...[By SSS congruence rule]
Similarly, ΔDEF ≅ ΔEDB
And ΔDEF ≅ ΔCFE
So, ΔABC is divided into four congruent triangles.
Hence proved.
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Remark: Figure is incorrect in Question
