Question

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

━━━━━━━━━━━━━━━━━━━━━━━━━

$\huge\underline\bold\orange{Question}$

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.

━━━━━━━━━━━━━━━━━━━━━━━━━

$\huge\underline\bold\orange{Solution}$

Let ∆ABC be a given triangle in which E and F are the midpoints of AB and AC respectively, Let AL be a line segment drawn from vertex A to the base BC, meeting BC at L and EF at M.

We have to show that AM = ML.

Through A, draw PAQ || BC.

In ∆ABC; E and F Bheem the midpoints of AB and AC respectively, we have EF || BC.

Now, PAQ, EF and BC are three parallel lines such that the intercepts AE and EB made by them on transversal AEB are equal.

∴ the intercepts AM and ML made by them on transversal AML must be equal

Hence, AM = ML.

━━━━━━━━━━━━━━━━━━━━━━━━━

Answer 2

Answer:

An=Ml answer

Hope it will help you