Answer:
Given : A Δ ABC, in which L M is a line segment , L being mid point of AB and M being mid point of AC.
To prove: L M║BC
Construction: Draw line segment L M , such that L ,and M are mid points of AB and AC respectively.
Proof: In Δ ABC
L, M are mid points of AB and AC respectively.
So, AL = 2 AB, and AM=2 AC
Taking Δ A L M and Δ ABC into consideration
[tex]\frac{AL}{AB}=\frac{AM}{AC}=\frac{1}{2}[/tex] →[Given]
∠A is common.
Δ A L M is similar to Δ ABC→[By side angle side axiom]
∠AL M= ∠ABC and ∠AM L=∠ A CB →[when triangles are similar corresponding angles are equal.]
But (∠ A L M and ∠ ABC) and ( ∠AM Land ∠ A CB) are pair of corresponding angles.
So, L M ║ BC →[When corresponding angles are equal lines are parallel]
