... follows: On a short extension of AB, drive a stake at C, and another at any convenient point P. Lay off PD = CP and PE = BP. The extensions of AP and DE intersect at F. Now prove that EF is the same length as AB. CUMULATIVE TEST 1 1.
|Title||:||The Scribner Plane Geometry|
|Author||:||Arthur F. Leary, Carl Nathaniel Shuster|
|Publisher||:||New York : Scribner - 1955|