For, (E. 32. 1. Cor. 1.) the four angles of the figure ADBC are together equal to four right; and, by the hypothesis, the four are the double of the two, DAC, ACB; it is manifest, ..., that the two DAC, ACB are together equal to two right;.. (E. 28. 1.) AD is parallel to CB: And, in the same manner, AC may be shewn to be parallel to DB; .. the figure ADBC is a parallelogram.

PROP. XXXVII.

51. PROBLEM. In a given square to inscribe an equilateral triangle, having one of its angular 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 given 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.

จ

Trisect (S. 20. 1.) the right and BF; bisect (E. 9. 1.) the BG and BH, meeting AD and respectively; and join G, H: equilateral.

ABC, by BE

ABE, CBF by DC in G and H, The A GBH is

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. 30. def. 1.) the A=2C, and the side AB, of the A ABG, is equal to the side CB, of the ▲ CBH; ..(E. 26. 1.) BG=BH; BH; .. (E. 5. 1.) the 2 BGH, =▲ BHG; also, (constr. and E. 8. 1.) the GBD = HBD; and BK is common to the two BKG, BKH; .. (E. 26. 1.) the BKG= 2BKH; .. (E. 10. def. 1.) each of these is a right ; .. (E. 32.1.) ≤ KGB+ ≤ GBK = a right ▲ = ▲ GBH + 22 ABG = ≤ GBH + ▲ ABE (constr.); but, since (constr.) ABG+ 2 CBH EBF, add to each of these equals the EBG, FBH, and the GBK ABE;

=

▲ GBH 24 ABE; .. the and it has been shewn that 4 KGB+ 4 GBK= 2 GBH+ABE; .. 4 KGB 2 GBH; ... HG= GB=BH ; i. e. the ▲ GBH is equilateral.

PROF. 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 pro. duced, two angles that are equal each of them to the vertical angle, they shall cut off equal segments from the base produced.

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 ABD, ACE, each equal to the vertical / BAC of the A, and let DF and EG, drawn from D and E, make with BC produced the DFB, EGC, each also equal to the < BAC: Then shall FB-GC.

L

For, from the point A draw (S. 25. 1.) AH and AK making with BC the AHB, AKC each equal to the DFB, or BAC, or CGE: And, since (E. 13. 1.) ▲ ABH+ 2 ABD+ 2 DBF=two right

= 2 DBF + 2 BFD + 2 FDB (E. 32. 1.), and that (constr.) ≤ ABD = 2 BFD, .. 2 ABH = 2 FDB; but, (constr.) AHB = ▲ DFB, and the side AB of the A AHB is equal to the side DB of the ▲ DFB; ... (E. 26. 1.) FB = AH: And in

be equal

2 AKC, (E. 6. 1.)

the same manner GC may be shewn to to AK; but since (constr.) the 2 AHB .. (E. 13. 1.) the 2 AHK = 2 AKH; .·. 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, 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 ABC, BCD, CDE, are together less than

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

For, join B, D: And (E. 32. 1.) ≤ DBC +2 BCD+4CDB=two right; if, ..., these three Ø be taken from the three given L, which (hyp.) are less than four right, there will remain the two

ABD, EDB, together less than two right; .. (E. 12. 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 ▲, right-angled

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

Trisect (S. 33. 1.) the hypotenuse C, in the points D and E; from D and E draw (E. 11. 1.) DF and EG 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 (hyp. and E. 5. 1.)

A is a right-angle, and that ▲ B=≤ C, ... (E. 32. 1.) <B