Geometry Problem
Click the figure below to see the complete problem 512 about Triangle, Three Squares, Concurrency.
See also:
Complete Problem 512
Level: High School, SAT Prep, College geometry
Thursday, August 26, 2010
Problem 512: Triangle, Three Squares, Concurrency
See also:
Complete Problem 512
Level: High School, SAT Prep, College geometry
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Let P is the point of intersection of MB and CN. We will show that 3 points K,A and P are collinear.
ReplyDeleteTri ABC similar to tri. KMN with ratio BC/MN=AC/KN=PC/PN
Note that angle (PCA)=angle (PNK) (corresponding angle)
So tri. PCA similar to tri. PNK (case SAS)
And angle(PAC)=angle(PKN) and 3 points P,A and K are collinear
Peter Tran
To Peter
ReplyDeleteIt is not enough, two tr to be similar just by congruence of an angle
Can you be more clearly, please
To c.t.e.o
ReplyDeleteConsider 2 triangles PAC and PKN
Angle PCA= Angle PNK ( corresponding angles)
and PC/PN=BC/MN ( BC//MN and tri PBC similar to tri PMN))
but BC/MN= AC/KN ( Tri. ABC similar to KMN)
so PC/PN=AC/KN
so tri. PAC similar to tri. PKN per case SAS
Peter Tran