Question

Given: delta'ABC Prove: A midsegment of Delta ABC is parallel to a side of delta ABC

Answer 1

S.A for Edmentum and Plato

Like and Rate

Answer 2

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

`(AL)/(AB)=(AM)/(AC)=(1)/(2)` →[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]