points upon one of the angular points of the square, and its two remaining angular points one in each of two adjacent sides of the square.

Let ABCD be a square: It is required to

inscribe in it an equilateral triangle, having one of its angular points upon the angular point B of the square.

Trisect (Supp. xx. 1.) the right ABC, by BE and BF; bisect (E. Ix. 1.) the 48 ABE, CBF by BG and BH, meeting AD and DC in G and H, respectively; and join G, H: The A GBH is equilateral.

For, join B, D, and let BD meet GH in K: Then, it is manifest from the construction, that the ▲ ABG= ▲ CBH; also, (hyp. and E. xxx. Def. 1.) the AC, and the side AB, of the AABG, is equal to the side CB, of the ACBH; therefore (E. XXVI. 1.) BG=BH; therefore (E. v. 1.) the ▲ BGH = ▲ BHG; also, (constr. and E. VIII. 1.) the GBD= ¿HBD; and BK is common to the two As BKG, BKH; therefore (E. xxvI. 1.) the <BKG = ¿BKH; therefore (E. x. Def. 1.) each of these angles is a right angle; therefore (E. XXXII. 1.) < KGB+ 4 GBK=a right <= 4 GBH÷24 ABG

= 2 GBH+LABE (constr.); but, since (constr.) LABG + ▲ CBH = 4EBF, add to each of these equals the 4s EBG, FBH, and the 4GBH= 2 LABE; therefore the GBK=LABE; and it has been shewn that KGB+4GBK=4GBH+ LABE; .. LKGB=4GBH; .. HG=GB=BH; i. e. the GBH is equilateral.

PROP. XXXVIII.

52. THEOREM. If, at the extremities of the base of a given triangle, two straight lines be drawn, both above the base, and each of them equal to the adjacent side, and making with it an angle equal to the vertical angle of the triangle; then, if two straight lines, let fall from the extremities of the two so drawn, make, with the base produced, two angles that are equal each of them to the vertical angle, they shall cut off equal segments from the base produced.

D

From the extremities B, C, of the base BC of the

given ABC, let BD be drawn equal to the adjacent side AB, and CE equal to the adjacent side AC, making the 48 ABD, ACE, each equal to the vertical <BAC of the triangle, and let DF and EG, drawn from D and E, make with BC produced

the 48 DFB, EGC, each also equal to the BAC: Then shall FB = GC.

For, from the point A draw (Supp. xxv. 1.) AH and AK making with BC the 48 AHB, AKC each equal to the DFB, or BAC, or CGE: And, since (E. xii. 1.) ▲ ABH+2ABD+2 DBF= two right angles =2DBF+2BFD+2FDB (E. xxxII. 1.), and that (constr.) LABD=2BFD; .. LABH=2FDB; but, (constr.) LAHB=LDFB, and the side AB of the AAHB is equal to the side DB of the DFB; therefore (E. XXVI. 1.) FB=AH: And in the same manner GC may be shewn to be equal to AK; but since (constr.) the LAHB=LAKC, therefore (E. xiii. 1.) the LAHK LAKH; therefore (E. VI. 1.) AH=AK; and FB was shewn to be equal to AH, and GC to AK; .. FB=GC.

53. COR. If the vertical <BAC be a right angle, the two straight lines AH and AK coincide; and the segments FB, GC are equal each of them to the perpendicular drawn from A to the base BC: In this case, also, DF= BK, and EG=CK.

PROP. XXXIX.

54. THEOREM. If four straight lines cut each other, without including space, but so as to make three internal angles, towards the same parts, which together are less than four right angles, the two lines, which are not joined, shall meet, if produced far enough.

Let the four straight lines AB, BC, CD, DE, cut one another, without enclosing space, so that

the 48 ABC, BCD, CDE, are together less than

four right angles; Then shall BA and DE meet, if they are produced far enough.

For, join B, D: And (E. XXXII. 1.) ▲ DBC+ LBCD + CDB = two right angles; if, therefore, these three angles be taken from the three given angles, which (hyp.) are less than four right angles; there will remain the two 4s ABD, EDB, together less than two right angles; therefore (E. xii. Axiom 1.) BA and DE will meet if they be continually produced.

PROP. XL.

55. PROBLEM. To inscribe a square in a given right-angled isosceles triangle.

Let ABC be the given isosceles triangle, right

angled at A: It is required to inscribe a square in the ABC.

Trisect (Supp. xxxiii. 1.) the hypotenuse AC, in

the points D and E; from D and E draw (E. xI. 1.) DF and EG perpendicular to BC, meeting the sides AB and AC in F, and G, respectively; and join F, G: The inscribed figure FDEG is a square.

For, since the LA is a right angle, and that (hyp. and E. v. 1.) <B= <C; therefore (E. xxxii. 1.) ≤B is half a right angle; but (constr.) the D is a right angle; therefore the DFB is half a right angle, and is, therefore, equal to the FBD; therefore (E. vi. 1.) DF=BD; but (constr.) BD=DE; .. DF=DE; and, in the same manner, it may be shewn that EG= DE; .. DF=DE=EG.

=

Again, since (constr.) the 4s D and E are right angles, therefore (E. xxvIII. 1.) DF is parallel to EG; and it has been shewn that DF EG; therefore (E. XXXIII. 1.) FG is equal and parallel to DE; therefore (E. xxix. 1.) the figure FDEG has all its angles right angles; and it is equilateral; therefore (E. xxx. Def. 1.) it is a square.

PROP. XLI.

56. PROBLEM. To find a point, in either of the equal sides of a given isosceles triangle, from which, if a straight line be drawn, perpendicular to that side, so as to meet the other side produced, it shall be equal to the base of the triangle.

Let ABC be the given isosceles triangle: It is required to find, in either of the two equal sides, as AB, a point from which if a perpendicular be drawn to AB and produced to meet AC, produced, it shall be equal to the base BC.

Draw (E. XI. 1.) from B, BD perpendicular to