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
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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