# Triangles, Squares and Areas from Temple Geometry

The problem presented by the applet below has been plucked from a collection of Temple Geometry Problems. Down below, I'll give three solutions, the first being drawn from another collection.

### Problem

Five squares are arranged as in the applet. Show that the area of triangle KMN equals the area of the square BEKH.

2 January 2016, Created with GeoGebra

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Copyright © 1996-2018 Alexander Bogomolny

Five squares are arranged as in the applet. Show that the area of triangle KMN equals the area of the square BEKH.

First of all note that angles ABE and CBJ being complementary, right triangle ABE and BCH are congruent. Therefore, if we denote AB as a and BC as b,

CY = CH = AB = a, and

XA = AE = BC = b.

### Solution 1

The construction needed for the first solution is hinted to in the applet.

Observe that triangles EFK and GHK are congruent to ABE and BCH. Thus, also

EF = GK = a, and

FK = GH = b.

This makes DF = 2a and GI = 2b. Now, triangles DFK and JKM are congruent, as are triangles GIK and KLN, so that

JK = FK = b,

JM = DF = 2a,

KL = GK = a,

LN = GI = 2b.

We see that

Area(KMN) | = Area(JMNL) - Area(JKM) - Area(KLN) |

= (a + b)(2a + 2b)/2 - 2a·b/2 - a·2b/2 | |

= (a + b)² - 2ab | |

= a² + b² | |

= Area(BEKH). |

### Solution 2

The only additional construction needed is line DI, which is not actually shown. The diagram presents several Vecten configurations, from which we conclude that triangles ABE, DEK, BCH, KHI have equal areas, as are triangles KMN and DIK.

Area(KMN) | = Area(DIK) |

= Area(DXYIK) - Area(DXYI) | |

= (a² + b² + 4ab/2 + a² + b²) - (2a + 2b)(a + b)/2 | |

= 2a² + 2b² + 2ab - a² - 2ab - b² | |

= a² + b² | |

= Area(BEKH). |

### Solution 3

Compared to the first solution, the second one draws very little on algebra. The third solution suggested by Nathan Bowler continues the progression: on the surface it does not use any algebra at all. I placed it on a separate page.

### References

- H. Fukagawa, D. Pedoe,
*Japanese Temple Geometry Problems*, The Charles Babbage Research Center, Winnipeg, 1989, #4.2.4 - J. Konhauser, D. Velleman, S. Wagon,
*Which Way Did the Bicycle Go?*, MAA, 1996, #50

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