Answer to Question #180601 in Geometry for Jas

Question

in a parallelogram ABCD,E,F,G and H are points on AB, BC, CD and DA, respectively such that EG is parallel to BC and FH is parallel to AB. Let P be the point of intersection of EG and FH. Show that the lines AF, CE and DP are concurrent.

Answer 1

Refer to the image attached.

Theory:

Converse Ceva's Theorem

For a triangle ABC and P,Q,R be the point on BC, CA and AB respectively.

then the line passing through P,Q,R are concurrent if and only if

$(\frac{AR}{RB})\cdot(\frac{BP}{PC})\cdot(\frac{CQ}{QA})= 1$

Solution:

Given ABCD is a parallelogram.

Consider E,F,G,H are mid points and lie on AB, BC, CD and DA respectively.

Now,

In triangle ABC

length of AE= length of BE ....(E is midpoint of AB)

length of BF= length of CF ....(F is midpoint of BC)

length of AP= length of PC ....(P is center and diagonals of parallelogram bisect each other.)

$(\frac{BE}{AE})\cdot(\frac{AP}{PC})\cdot(\frac{CF}{BF})=(\frac{AE}{AE})\cdot(\frac{PC}{PC})\cdot(\frac{BF}{BF})\newline \quad\newline \therefore(\frac{BE}{AE})\cdot(\frac{AP}{PC})\cdot(\frac{CF}{BF})=1$

Using Ceva's Theorem

The lines AF, CE and BP(i.e DP) are concurrent.

Hence Proved.