Question

prove that any straight line drawn from the vertex of a triangle to the base is bisected by straight line which joins the middle points of other sides of a triangle​

Answer 1

Hey,

I have mentioned the answer in the above picture.
Since you haven't specified wihich triangle. I have proven your thesis in the context of an equilateral and equiangular triangle.

Hope this helps. You can clarify any further doubts.

☺️

Answer 2

Proved below.

Step-by-step explanation:

Given:

In Δ ABC, AE is line from A to BC.

Also D, E, F are the mid points of AB, BC, AC.

Let Q be point where DF intersect AE.

To prove:

DF bisects AE that is Q is the mid point of AE.

Proof:

Since DF is line joining mid points ,

therefore, DF║BC,

$DF = \frac{1}{2} BC$ (Mid point theorem, the line segment joining the mid points of two sides of a triangle is parallel to the third side and is half of it. )

Now in Δ ABE,

D is mid point of AB

DQ║BE (parts of parallel lines are parallel )

So, Q is the mid point of AE (converse of Mid point theorem)

Therefore, DF bisects AE.

Hence proved