Suppose we were to cut the triangle _d e f_ out of the page and place it upon the triangle _a b c_, so that the line _e f_ shall fall upon the line _b c_, with the point _e_ upon the point _b_. Because the line _e f_ is equal to the line _b c_, upon what point will the point _f_ fall? Because the angle _e_ is equal to the angle _b_, where will the line _e d_ fall? Because the angle _f_ is equal to the angle _c_, where will the line _d f_ fall? Then, if the line _d e_ falls upon the line _a b_ and the line _d f_ upon the line _a c_, where will the point _d_ fall? Now because the three sides of the triangle _d e f_ exactly fall upon the three sides of the triangle _a b c_, we say _the two magnitudes coincide throughout their whole extent, and are therefore equal_. Suppose the angle _e_ were greater than the angle _b_, would the line _e d_ fall within or without the triangle? If it were less, where would the line fall? Why does the line _d e_ fall exactly upon the line _a b_? [Illustration] DEMONSTRATION. We wish to prove that, _If two triangles have two angles, and the included side of the one equal to two angles and the included side of the other, each to each, the two triangles are equal to each other in all respects._ Let the triangles _a b c_ and _d e f_ have the angle _b_ of the one equal to the angle _e_ of the other; the angle _c_ of the one equal to the angle _f_ of the other; and the included side _b c_ of the one equal to the included side _e f_ of the other, each to each; then will the two triangles be equal in all their parts. For place the triangle _d e f_ upon the triangle _a b c_, so that the line _e f_ shall fall upon the line _b c_, with the point _e_ upon the point _b_. Because the line _e f_ is equal to the line _b c_ the point _f_ will fall upon the point _c_. Because the angle _e_ is equal to the angle _b_, the line _e d_ will fall upon the line _b a_, and the point _d_ will be somewhere in the line _b a_. Because the angle _f_ is equal to the angle _c_, the line _f d_ will fall upon the line _c a_, and the point _d_ will be somewhere in the line _c a_. Then, because the point _d_ is in the two lines, _b a_ and _c a_, it must be in their intersection, or upon the point _a_. And, as the two triangles coincide throughout their whole extent, they are equal in all their parts. That is, the angle _a_ is found to be equal to the angle _d_; the side _b a_ to the side _e d_; the side _c a_ to the side _f d_; and the area of the triangle _a b c_ to the area of the triangle _d e f_. [Illustration] PROPOSITION XV. THEOREM. DEMONSTRATION. We wish to prove that _The opposite sides of any parallelogram are equal._