Online Geometry theorems, problems, solutions, and related topics.
Geometry ProblemClick the figure below to see the complete problem 512 about Triangle, Three Squares, Concurrency.
Let P is the point of intersection of MB and CN. We will show that 3 points K,A and P are collinear.Tri 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 collinearPeter Tran
To PeterIt is not enough, two tr to be similar just by congruence of an angleCan you be more clearly, please
To c.t.e.oConsider 2 triangles PAC and PKNAngle 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/KNso tri. PAC similar to tri. PKN per case SASPeter Tran