Anonymous

# Geometry help needed? Make my day?

ABC triangle rectangle in B. AB=6 cm and BC=8cm. N is the middle of [BC]. E is the point of the line (BC) exterieur from the segment [BC] and BE=2cm.

The bisector [Cx) of the angle ACE cuts [AE] at M and [AB] at P.

1) What represents P for the triangle AEC? Justified.

2) (EP) cut [AC] at K. Show that AKE is a triangle rectangle in K.

3) [TM] is symmestric about point E. Show that B is the centre of gravity of the triangle MNT.

4) (MB) cut [TN] at S. Show that (ES) and (MN) are parallel.

Feel free to use any methods and write steps so it's faster for you.

Thanks guys

Relevance

1) P is the orthocenter of AEC (point where altitudes intersect)

explanation: In triangle ABC, AC^2 =AB^2 + BC^2 = 36+64=100,

so AC = 10

Also, CE = CB+BE = 8+ 2 = 10.

As AC = CE, the triangle ACE is thus isoceles.

In this triangle,Cx is the bisector, and because it's isoceles, Cx is also an altitude.

Both Cx and AB are altitudes in triangle ACE, and they intersect at P, so P is the orthocenter.

2) We previously said that P is the orthocenter of AEC, so all lines passing from a vertex through P will be an altitude. EK is thus the third altitude of the triangle and angle AKE = 90 degrees. So AKE is a triangle rectangle in K.

3) The center of gravity is the intersection of the medians in a triangle.

as TM is symmetric about E, ME = ET, and NE is a median, so the center of gravity lies on that line.

One property of the center of gravity is that, on a given median, it lies 2/3 of the way from the vertex, and 1/3 from the base.

We have that BE = 2 and BN = 4 , so EN = 6

BE / NE = 2/6 = 1/3.

As point B verifies this property, it is the center of gravity

4) Since MS passes through B, it is a median in triangle MNT, and NS = ST.

TS /TN = TS/2TS = 1/2

TE/TM = TE/2TE = 1/2

This ratio verifies the inverse Thales theorem in a triangle, where if the ratios are equal, then the lines passing through the points forming these ratios are parallel. Thus ES and MN are parallel