D, E , F are the midpoints of the sides BC, CA and AB respectively of triangle ABC, then triangle DEF is congruent to triangles
1 point
(a) ABC
(b) AEF
(c) BFD, CDE
(d) AFE, BFD, CDE
in
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Step-by-step explanation:
ANSWER
Given: In ΔABC, D,E and F are midpoints of sides AB,BC and CA respectively.
BC=EC
Recall that the line joining the midpoints of two sides of a triangle is parallel to third side and half of it.
Therefore,we have:
DF=d
2
1
BC
⇒
BC
DF
=
2
1
....(1)
AC
DE
=
2
1
....(2) and
AB
EF
=
2
1
....(3)
From (1), (2) and (3) we have
But if in two triangles, sides of one triangle are proportional to the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar
Therefore, ΔABC∼ΔEDF [By SSS similarity theorem]
Hence area of ΔABC: area of ΔDEF=4:1
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