Geometry Problem. Post your solution in the comment box below.

Level: Mathematics Education, High School, Honors Geometry, College.

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## Friday, March 15, 2019

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

Labels:
collinear,
concurrent,
geometry problem,
line,
point,
sketch,
two squares

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Suppouse : a)AG and BH meet on P b) DF and CE meet on P'. Draw PT, P'T' ꓕ AH

ReplyDelete=> 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

Let u=AB and v= GH

ReplyDeleteTriangle 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

Triangle ADP1 simillar to GFP1

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

To Peter

DeleteP1A/P1G ≠ PA/PG ?

yes, line 6 of my solution will be:

Delete"So P1 is a point locate on AG such that P1A/P1G= u/v => P1 coincide to P"

Peter

But A, P1, G are not collinear

DeleteLet AG and BH meet at P

ReplyDeleteLet 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