An interesting variation of the ordinary proof is made by placing a trapezoid _T'_, congruent to _T_, in the position here shown. The parallelogram formed equals _a_(_b_ + _b'_), and therefore _T_ = _a_ . (_b_ + _b'_)/2. The proposition should be discussed for the case _b_ = _b'_, when it reduces to the one about the area of a parallelogram. If _b'_= 0, the trapezoid reduces to a triangle, and _T_ = _a_ . _b_/2. This proposition is the basis of the theory of land surveying, a piece of land being, for purposes of measurement, divided into trapezoids and triangles, the latter being, as we have seen, a kind of special trapezoid. The proposition is not in Euclid, but is given by Proclus in the fifth century. The term "isosceles trapezoid" is used to mean a trapezoid with two opposite sides equal, but not parallel. The area of such a figure was incorrectly given by the Ahmes papyrus as 1/2(_b_ + _b'_)_s_, where _s_ is one of the equal sides. This amounts to taking _s_ = _a_. The proposition is particularly important in the surveying of an irregular field such as is found in hilly districts. It is customary to consider the field as a polygon, and to draw a meridian line, letting fall perpendiculars upon it from the vertices, thus forming triangles and trapezoids that can easily be measured. An older plan, but one better suited to the use of pupils who may be working only with the tape, is given on page 99. THEOREM. _The areas of two triangles which have an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles._ This proposition may be omitted as far as its use in plane geometry is concerned, for we can prove the next proposition here given without using it. In solid geometry it is used only in a proposition relating to the volumes of two triangular pyramids having a common trihedral angle, and this is usually omitted. But the theorem is so simple that it takes but little time, and it adds greatly to the student's appreciation of similar triangles. It not only simplifies the next one here given, but teachers can at once deduce the latter from it as a special case by asking to what it reduces if a second angle of one triangle is also equal to a second angle of the other triangle. It is helpful to give numerical values to the sides of a few triangles having such equal angles, and to find the numerical ratio of the areas. THEOREM. _The areas of two similar triangles are to each other as the squares on any two corresponding sides._ [Illustration] This may be proved independently of the preceding proposition by drawing the altitudes _p_ and _p'_. Then [triangle]_ABC_/[triangle]_A'B'C'_ = _cp_/_c'p'_. But _c_/_c'_ = _p_/_p'_, by similar triangles. [therefore] [triangle]_ABC_/[triangle]_A'B'C'_ = _c_^2/_c'_^2, and so for other sides.