and IG is common to the two triangles EIG, FIG, and the angles at I are equal, being right angles; therefore (I. 4) the triangles are equal in every respect, and therefore EG = GF; also angle GEI= GFI. (1.) Again, because angle GEI = GFI, therefore KFI, which exceeds GFI, is greater than GEI, and consequently in the triangle KEF the angle KFE is greater than the angle KEF, therefore (I. 19) the side KE is greater than KF; or K is unequally distant from E and F. (2.) If KI were joined, the latter part is easily proved by (I. 24), for EI and IK are Р K E I F L equal to FI and IK; but the angle EIK is greater than the angle FIK, therefore the base EK is greater than the base FK. (3.) Or, since it has been proved that EG: GF, to each add GK, then EG and GK, that is, EK = FG+ GK, but FG and GK are greater than FK (I. 20), therefore EK is greater than FK. EXERCISE III.-PROBLEM. = In a given straight line, to find a point equally distant from two given points. Let AB be the given line, and PQ the given points, to find a point, as C in AB, that shall be equidistant from the points P and Q. Join P and Q, bisect PQ in D, and draw CD perpendicular to PQ, and produce it to cut AB in C; and C is the required point. For in the two triangles CDP, CDQ, PD DQ, and CD is common, and the A angles CDP, CDQ are equal, being P B right; consequently (I. 4) the triangles are equal in all respects, and CP CQ; wherefore C is the required point. = It is evident, when the given points are on opposite sides of the given line, or when one of the points is in the given line, that the same construction and proof will apply as in the above case, when the points are on one side of the line. But when the two points are so situated that the line joining them is perpendicular to the given line, the problem is impossible (unless they happen to be equidistant from the line, when every point on the line would fulfil the condition); for the line bisecting perpendicularly the line joining them, would be parallel to the given line, and would therefore never meet it. EXERCISE IV.-THEOREM. If any line be drawn through the middle point of the line joining two given points, any two points in the former line that are equidistant from the middle point, are also equidistant from the two given points. Let MN be the given points, O the middle of MN, and XY any line through O; also let OX = OY, and join XM, YN, then will MX = NY. For in the triangles OMX, ONY, the sides OX, OM, in the one, are respec tively equal to OY, ON, in the other, X R M N IS Y It is evident that, in whatever direction the line XY is drawn, provided it passes through the point O, the same reasoning will apply. EXERCISE V.-THEOREM. Of all lines that can be drawn from a given point to a given line, the perpendicular upon it is the least; and of all others, that which is nearer to the perpendicular is less than the more remote; and only two equal lines can be drawn to it from that point, one upon each side of the perpendicular. Let AB be a given straight line, and P a point without it; to prove that, of all lines drawn from P to AB, PC the perpendicular is the least, and PD, which is nearer to PC, is less than PE, which is more remote, and PE is less than PA. Since PC is perpendicular to AB, PCD is a right angle; therefore (I. 17) PDC is less than a right angle; therefore, since the angle PDC is less than the angle A PCD, the side PC is less than PD E D C B (I. 19). Again, PDC being an acute angle, PDE is an obtuse angle (I. 13), and is therefore greater than PED; PE is greater than PD: in the same manner it can be proved that PA is greater than PE; hence PC is the least line that can be drawn from P to the line AB, and PD is less than PE, and PE is less than PA. Also, there can only be drawn two equal lines from P to AB, one on each side of the perpendicular PC; for make CB = CD, and join PB, then PC and CD are equal to PC and CB, and the contained angles PCD and PCB are also equal; therefore the base PD PB. But besides PB, no other line can be drawn equal to PD, for if it were nearer to the perpendicular it would be less, and if more remote it would be greater; hence there can only be drawn two equal lines from a point to a straight line, one on each side of the perpendicular. EXERCISE VI.-THEOREM. If from every point of a given line, the lines drawn to each of two given points on opposite sides of the line are equal; prove that the straight line joining the given points will be bisected by the given line at right angles. Let AB be a given line, and P, Q, two points on opposite sides of it, such that whatever point may be taken in AB, the lines drawn from it to P and Q shall be equal; to prove that AB bisects PQ, and cuts it at right angles. B Take the points B and C in AB, and join BP, BQ, and CP, CQ; then in the triangles PBC and QBC, PB and BC are equal to QB and BC, and the base PC is equal to the base QC (by hypothesis), therefore the angle PBC is equal to the angle QBC (I. 8). Again, in P the triangles PBD and QBD, PB = BQ, and BD is common, and the angle PBD was proved equal to the angle QBD, therefore the base PD is equal to the base QD, and the angle PDB = QDB (I. 8), and they are adjacent angles, therefore each of them is a right angle; hence AB bisects PQ, and cuts it at right angles. EXERCISE VII.-THEOREM. D In the figure, Euclid I. 5, if BG and CF meet in O, shew that OA bisects the angle BAC. Referring to the figure of (I. 5), in the triangles FBC and GCB, which are proved equal in every respect, the angle FCB, that is, OCB, is equal to the angle GBC, that is, OBC; hence in the triangle OBC, since the angle OBC is equal to the angle OCB, the side OB = OC (I. 6). Again, in the triangles BAO, CAO, the side BA is equal to AC, and AO common, and the base BO has been proved equal to the base CO; therefore the angle BAO is equal to CAO (I. 8), and the line AO bisects the angle BAC. EXERCISE VIII.-THEOREM. The difference between two sides of a triangle is less than the third side. Let ABC be a triangle; then the difference between any two of its sides, as AB and AC, is less than the third side BC. For of the two sides AB, AC, let AB be the greater, and from it cut off a part, AD AC the less, then (I. 20) AC and CB are A D B greater than AB or AD and DB; and if from these unequals the equals AC and AB be taken away, the remainder CB is greater than the remainder DB. When the sides AB, AC, happen to be equal, their difference is nothing, and the proposition is evident; it is also evident when the third side BC is equal to the greater side AB. EXERCISE IX.-THEOREM. If two isosceles triangles be constructed on opposite sides of the same base, the line joining their vertices bisects the common base, and each of the vertical angles. Let CAB and DAB be two isosceles triangles on opposite sides of the same base AB, it is required to prove that the line joining their vertices C and D bisects AB and each of the angles ACB and ADB. Since the triangles ACD and BCD have AC = CB, AD DB, and CD common, they are (I. 8) equal in all respects, and have the angles equal that = A B are opposite to the equal sides; therefore the angle ACD = BCD, and the angle ADC BDC. Again, in the triangles ACE and BCE, the two sides AC and CE are equal to the two sides BC and CE, and the contained angles ACE and BCE were proved equal, therefore (I. 4) the base AE is equal to BE, and the line AB, which is the common base, is bisected in E. EXERCISE X.-THEOREM. Every point in the line that bisects a given angle is equidistant from the sides of the angle. Let AB, AC, be the sides of an angle BAC, which is bisected by the line AD; then, if from any point O in AD, the perpendiculars OE, OF be drawn upon the sides, then is OE OF. = For in the triangles AOE, AOF, the angles at A are equal by construction, the angles at E and F are right angles, and the side AO is common to the two triangles; consequently (I. 26) they are equal in every respect, and therefore A OE OF. C E F EXERCISE XI.-THEOREM. If the alternate extremities of two equal and parallel lines be joined, the connecting lines bisect each other. Let MN, KL be two equal and parallel lines; then ML and KN, joining their alternate extremities, bisect each other in O. For in the two triangles MON, KOL, the vertical angles at O are equal (I. 15), the alternate angles at M and L (I. 29) are equal, and the sides KL, MN are also equal; therefore (I. K 26) the triangles are equal in every respect, and M L N consequently OK = ON, and OLOM; that is, ML and KN are bisected in O. EXERCISE XII.-THEOREM. If the vertical angle of an isosceles triangle be a right angle, each of the angles at the base is half a right angle. Let PQR be an isosceles triangle, having its vertical angle R a right angle; then each of the angles P and Q, at the base, is half a right angle. For the three angles P, Q, R, of the triangle PQR, are together equal to two right angles, of which R is a right angle; therefore the sum of the angles at P and Q is equal to a right angle; and since they are also equal (I. 5), therefore each of them is half a right P angle. EXERCISE XIII.-THEOREM. R S If a side of an isosceles triangle be produced beyond the vertex, the exterior angle is double of either of the angles at the base. Let PQR (fig. to Ex. 12) be an isosceles triangle, and let the side PR be produced beyond the vertex to S, then the exterior angle QRS is double of either of the angles at P or Q. For (I. 32) the exterior angle QRS is equal to the two interior opposite angles P and Q together; but these angles being equal (I. 5), their sum is double of either of them; consequently the exterior angle QRS is double of the angle at P or the angle at Q. EXERCISE XIV.-THEOREM. The middle point of the hypotenuse of a right-angled triangle is equally distant from each of the three angles. Let ABC be a right-angled triangle, having the right angle ACB, and let D be the middle point of the hypotenuse, to prove that AD, DC, and DB are equal. If DC be not equal to AD or DB, let DC be the greater; then since DC is greater than AD (I. 18), the angle CAD B |