Question

$\large \textbf {Maths \: Question}$

1) Prove that any line segment drawn from the vertex of a triangle to the base is bisected by the line segment joining the midpoints of the other side of the triangle.

Answer 1

Given : E and F are mid points on side AB and AC of triangle respectively. AL is line segment drawn from A to BC.

To Prove : AM = LM

Proof : By theorem we know that,

Line joining mid points of triangle are always parallel and half to that of base.

Therefore, EF || BC

In ∆ABL and ∆ACL,

That's why, EM || BL and MF || LC

Here E is mid point on AB and F is mid point on AC.

Hence by theorem mentioned above,

Point M is mid point on line segment AL.

Hence, AL = LM

Q.E.D

Answer 2

$\huge \underline \mathfrak {Solution:-}$

E and F are the midpoint of AB and AC respectively.

We know that, in a triangle, the line joining the midpoint of any two sides is parallel to the third side.

Here we can show that EF || BC.

Now, in ∆AMF and ∆ACL

angle AFM = angle ACL (corresponding angles)

angle AMF = angle ALC (corresponding angles)

So, ∆AMF ≈ ∆ACL

Hence,

$\frac{AF}{AC} = \frac{AM}{AL} \: \: \: \: \: .....(1)$

ACCORDING TO QUESTION

AF/AC = 1/2 (F is the midpoint of AC)

FROM EQ (1)

AM/AL = 1/2

AM/(AM+ML) = 1/2

2AM = AM + ML

AM = ML

THEREFORE, M IS THE MIDPOINT OF AL.