## Friday, March 15, 2019

### Geometry Problem 1425: Two Squares, Collinear Points, Concurrent Lines

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below.

1. Suppouse : a)AG and BH meet on P b) DF and CE meet on P'. Draw PT, P'T' ꓕ AH
=> From similarity of triangles AB/HG = AT/HT and DC/FE = T'D/T'E
=> AT/HT = T'D/T'E . This happen when P and P' coincide

2. Let u=AB and v= GH
Triangle ABP simillar to GPH ( case AA)
So P is a point locate on AG and BH such that PB/PH=PA/PG= u/v
Let P1 is the intersection of AG and FD
So P1 is a point locate on AG such that PA/PG= u/v => P1 coincide to P
Simillarly to other triangles with P2, P3 .
P2, and P3 with coincide to P

1. To Peter
P1A/P1G ≠ PA/PG ?

2. yes, line 6 of my solution will be:
"So P1 is a point locate on AG such that P1A/P1G= u/v => P1 coincide to P"
Peter

3. But A, P1, G are not collinear

3. Let AG and BH meet at P

Let AD = a, EF = b, AP = x and PG = y

From similar triangles ABP & GPH, x/a = y/b

In Tr.s APD & FPG, x/a = y/b and the included angles < DAP = GPF

So Tr.s APD & FPG are similar and < APD = < GPF

So < APF + < APD = < APF + < GPF = 180
Therefore points DPF are collinear

Similarly for points CPE

Sumith Peiris
Moratuwa
Sri Lanka