Proof. We have, now, two triangles DBC and ACB (DBC lying on A CB). D B In those triangles, DB is equal to AC (b), the two sides DB, B C are equal to the two sides A C, CB (B C being common), and angle D B C is equal to angle A CB (Given a). by Euc. I. 4, (c) The triangle D B C is equal to the triangle AC B. (d) But, the triangle DBC is less than the triangle ACB (self-evident). ... If AB is not equal to AC, the triangle DBC is both equal to and less than the triangle ACB; which is absurd. .. AB is not unequal to AC, that is, AB is equal to AC. Q. E. D. NOTE. The side of a triangle opposite to any par ticular angle is said to subtend that angle. EXERCISES.-I. Given.-A C equal to AD, and BC equal to BD (Fig. 1). Required. To prove that triangle A CB is equal to triangle A D B, without using Euc. I. 8. (Join CD, and use Euc. I. 5, and then I. 4) II. Given. The two isosceles triangles ACB, ADB (Fig. 2). Required. To prove that A E is equal to EB. (Apply Euc. I. 8 to the triangles CDA, CDB. Then Euc. I. 13 to the angles at D. Then apply Euc. I. 4 to the triangles A DE, BDE.) THEOREM (Euclid I. 7). Repeat.-The enunciation of Euc. I. 5 and Axiom 9. General Enunciation. On the same base, and on the same side of it, there cannot be two triangles having their sides which are terminated at one extremity of the base equal to one another, and likewise those sides which are terminated at the other extremity equal to one another. The same, in tabular form. GIVEN. Two triangles on the same base, and on the same side of the base. REQUIRED. To prove that— the two sides terminated at one end of the base cannot be equal if those terminated at the other end of the base are equal. Particular Enunciation (Case I). Given. The two triangles CAB, DAB, on the same base A B, and on the same side of it, and C A B having the vertex of each triangle without the other. Required. To prove that it is not possible that the two sides CA, DA should be equal, and, at the same time, the two sides DA, DB be equal (see explanation of this on page 74). Construction. Join CD. Proof. Let us suppose (a) that CA and D A are equal, and also that (b) CB and DB are equal. A B Then A CD is an isosceles triangle, having side A C equal to side A D (a). (c). by Euc. I. 5, angle A CD is equal to angle A D C. But angle A CD is greater than angle B CD. .. by Axiom 9, angle A D C is greater than angle BCD (c). (d) and angle BDC is still greater than angle BC D. Again, BDC is an isosceles triangle having side BD equal to side BC (b). (e).. by Euc. I. 5, angle BDC is equal to angle B CD, but we have proved that angle BDC is greater than angle B CD (d). .. if A C and AD are equal, and also BD and BC equal, then angle BDC is both greater than (d) and equal to (e) angle B CD; But this is impossible: and so it is impossible for AC and AD to be equal at the same time that BD and DC are equal. Q. E. D. Particular Enunciation (CASE II). Given. The two triangles CAB, DAB, on the same base A B, and on the same side of it; but having the vertex of one triangle within the other triangle. Required. To prove that A C and AD cannot be equal at the same time that DC and DB are equal. Construction. Join CD. (See Fig. on page 98.) Produce AC to E, and AD to F. Proof. Let us suppose, as before, (a) that A C, AD are equal, and Then A CD is an isosceles triangle having its equal sides A C, AD (a) produced to E and F. (c).. by Euc. I. 5, angle ECD is equal to angle F DC, But angle E CD is greater than angle B CD. BDC is an isosceles triangle, having H |