prove that the line segment joining the midpoint of the sides of a triangle divides it into four congruent triangles
Answers
Ello there!
------------------------------
Pre-Requisite Knowledge
@ Midpoint Theorem.
@ Converse of Midpoint Theorem.
-------------------------------
Answer
Given
D is the midpoint of AB
E is the midpoint of AC
F is the midpoint of BC
To prove
The triangles formed by the linesegment joining the midpoints are congruent.
Proof
In ΔABC
D and E are the midpoints of AB and AC,
∴ By Midpoint theorem,
DE║BC
DE║BF (Part of a parallel line is also parallel to the whole lines parallel line)
DE = BC
DE = BF (F is the midpoint)
In quadrilateral DEBF,
DE║BF and DE = BF
∴ DEBF is a parallelogram. (one pair of opposite sides are equal and parallel)
⇒ ΔDBF ≅ ΔDEF → 1
(Diagonal of a IIgm divides it into two congruent triangles)
Similarly,
AEFD is a parallelogram
⇒ ΔDAE ≅ ΔDFE → 2
(Diagonal of a IIgm divides it into two congruent triangles)
Similarly,
DECF is a parallelogram
⇒ ΔDFE ≅ ΔCFE → 3
(Diagonal of a IIgm divides it into two congruent triangles)
From 1, 2 and 3
ΔDBF ≅ ΔDEF ≅ ΔCFE ≅ ΔADE
Therefore, line segment joining the midpoint of the sides of a triangle divides it into four congruent triangles.
-------------------------------
Hope It Helps!
