0
$\begingroup$

Consider a triangle ABC. Let D is the mid point of side BC and let EF is a line segment parallel to BC and E , F lie on side AB , AC respectively.

How can one prove that if the median AD of triangle cuts the line seqment EF at P then P is mid point of EF as well. Ie in other words joining mid points of sides parallel to BC we get the median through A.

$\endgroup$
0
$\begingroup$

It follows from $$\frac{EP}{BD} = \frac{AE}{AB} = \frac{AF}{AC} = \frac{FP}{CD} \ \ \ \text{and } \ \ \ BD = CD \implies EP = FP$$ Thus $P$ is the midpoint of $EF$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.