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

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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