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 triangle​

Answer 1

$\bf\underline{\underline{\pink{Question:-}}}$

✭ 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 triangle

$\bf\underline{\underline{\blue{Given:-}}}$

✭ In △ABC; E and F being the midpoints of AB and AC respectively, we have EF || BC.

$\bf\underline{\underline{\red{ToProve:-}}}$

✭ AM = ML

$\bf\underline{\underline{\green{Solution:-}}}$

Let △ABC be a given triangle in which E and F aadat remix point of AB and AC respectively. Let AL b 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, drawvPAQ || BC.

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

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

∴ the intercepts AM and ML made by one transversal AML must be equal.

Hence, AM = ML