Geometry Problem

Click the figure below to see the complete problem 512 about Triangle, Three Squares, Concurrency.

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Complete Problem 512

Level: High School, SAT Prep, College geometry

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## Thursday, August 26, 2010

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Problem 512: Triangle, Three Squares, Concurrency

See also:

Complete Problem 512

Level: High School, SAT Prep, College geometry

Online Geometry theorems, problems, solutions, and related topics.

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

Labels:
concurrent,
square,
triangle

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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