To inscribe a square in a giten triangle. Analysis. Let ABC be the given triangle, of which the base BC, and the perpendicular AD are given.

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Let FGHK be the required inscribed square.

Then BHG, BDA are similar triangles,
and GH is to GB, as AD is to AB,

but GF is equal to GH;

therefore GF is to GB, as AD is to AB. Let BF be joined and produced to meet a line drawn from A parallel to the base B C in the point E.

Then the triangles BGF, BAE are similar,

and AE is to AB, as GF is to GB,
but GF is to GB, as AD is to AB;
wherefore AE is to AB, as AD is to AB;

hence AE is equal to AD. Synthesis. Through the vertex A, draw AE parallel to BC the base of the triangle,

make A E equal to AD,

join EB cutting AC in F, through F, draw FG parallel to BC, and FK parallel to AD;

also through G draw GH parallel to AD.

Then GHKF is the square required. The different cases may be considered when the triangle is equilateral, scalene, or isosceles, and when each side is taken as the base.

PROPOSITION II. THEOREM. If from the extremities of any diameter of a given circle, perpendiculars be drawn to any chord of the circle, they shall meet the chord, or the chord produced in two points which are equidistant from the center.

First, let the chord CD intersect the diameter AB in L, but not at right angles ; and from A, B, let AE, BF be drawn perpendicular to CD. Then the points F, E are equidistant from the center of the chord CD.

Join EB, and from I the center of the circle, draw IG perpendiular to CD, and produce it to meet EB in H.

Then IG bisects CD in G; (111. 2.) and IG, AE being both perpendicular to CD, are parallel. (I. 29.)

Therefore BI is to BH, as IA is to HE; (VI. 2.)

and BH is to FG, as HE is to GE: therefore BI is to FG, as IA is to GE;

but BI is equal to IA;

therefore FG is equal to GE.

It is also manifest that DE is equal to CF. When the chord does not intersect the diameter, the perpendiculars intersect the chord produced.

PROPOSITION III. THEOREM. If two diagonals of a regular pentagon be drawn to cut one another, the greater segments will be equal to the side of the pentagon, and the diagonals will cut one another in extreme and mean ratio.

Let the diagonals AC, BE be drawn from the extremities of the side AB of the regular pentagon ABCDE, and intersect each other in the point H.

Then BE and AC are cut in extreme and mean ratio in H, and the greater segment of each is equal to the side of the pentagon.

Let the circle ABCDE be described about the pentagon. (IV. 14.)
Because EA, AB are equal to AB, BC, and they contain equal

therefore the base EB is equal to the base AC, (1. 4.)

and the triangle EAB is equal to the triangle CBA, and the remaining angles will be equal to the remaining angles, each to each, to which the equal sides are opposite.


Therefore the angle BAC is equal to the angle ABE; and the angle AHE is double of the angle BĂH, (1. 32.) but the angle EAC is also double of the angle BAC, (VI. 33.)

therefore the angle HAE is equal to AHE,
and consequently HE is equal to EA, (1. 6.) or to AB.

“And because B A is equal to ,
the angle ABE is equal to the angle AEB;

but the angle ABE has been proved equal to BAH:
therefore the angle BEA is equal to the angle BAH:

and ABE is common to the two triangles ABE, ABH; therefore the remaining angle BAE is equal to the remaining

angle AHB; and consequently the triangles ABE, ABH are equiangular; therefore EB is to BA, as AB to BH: but BA is equal to EH,

therefore EB is to EH, as EH is to BH, but BE is greater than EH; therefore EH is greater than HB;

therefore BE has been cut in extreme and mean ratio in H. Similarly, it may be shewn, that AC has also been cut in extreme and mean ratio in H, and that the greater segment of it CH is equal to the side of the pentagon.

PROPOSITION IV. PROBLEM. Divide a given arc of a circle into two parts which shall have their chords in a given ratio.

Analysis. Let A, B be the two given points in the circumference of the circle, and C the point required to be found, such that when the chords AC and B C are joined, the lines AC and BC shall have to one another the ratio of E to F.


Draw CD touching the circle in C;

join AB and produce it to meet CD in D. Since the angle BAC is equal to the angle BCD, (III. 32.) and the angle CDB is common to the two triangles DBC, DAC; therefore the third angle CBD in one, is equal to the third angle DCA in the other, and the triangles are similar,

therefore AD is to DC, as DC is to DB; (VI. 4.) hence also the square on AD is to the square on DC, as AD is to BD. (vi. 20. Cor.)

But AD is to AC, as DC is to CB, (VI. 4.)

and AD is to DC, as AC to CB, (v. 16.)" also the square on AD is to the square on DC, as the square on AC

is to the square on CB; but the square on AD is to the square on DC, as AD is to DB: wherefore the square on AC is to the square on CB, as AD is to BD;

but A C is to CB, as E is to F, (constr.) therefore AD is to DB as the square on E is to the square on F.

Hence the ratio of AD to DB is given, and AB is given in magnitude, because the points A, B in the cir

cumference of the circle are given.

Wherefore also the ratio of AD to AB is given, and also the mag

nitude of AD. Synthesis. Join AB and produce it to D, so that AD shall be to BD, as the square on E to the square on F.

From D draw DC to touch the circle in C, and join CB, CA. Since AD is to DB, as the square on E is to the square on F, (constr.)

and AD is to DB, as the square on AC is to the square on BC; therefore the square on A C is to the square on BC, as the square on E is to the square on F,

and AC is to BC, as E is to F.

PROPOSITION V. PROBLEM. A, B, C are given points. It is required to draw through any other point in the same plane with A, B, and C, a straight line, such that the sum of its distances from two of the given points, may be equal to its distance from the third.

Analysis. Suppose F the point required, such that the line XFH being drawn through any other point X, and AD, BE, CH perpendiculars on XFH, the sum of BE and CH is equal to AD.

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в Join AB, BC, CA, then ABC is a triangle. Draw AG to bisect the base B C in G, and draw GK perpendicular to EF.

Then since BC is bisected in G,
the sum of the perpendiculars CH, BE is double of GK;

but CH and BE are equal to AD, (hyp.)
therefore AD must be double of GK;

but since AD is parallel to GK,
the triangles ADF, GKF are similar,

therefore AD is to AF, as GK is to GF; but AD is double of GK, therefore AF is double of GF; and consequently, GF is one-third of AG the line drawn from the vertex of the triangle to the bisection of the base. But AG is a line given in magnitude and position,

therefore the point F is determined. Synthesis. Join AB, AC, BC, and bisect the base BC of the triangle ABC in G; join AG and take GF equal to one-third of GA;

the line drawn through X and F will be the line required. It is also obvious, that while the relative position of the points A, B, C, remains the same, the point Fremains the same, wherever the point X may be. The point X may therefore coincide with the point F, and when this is the case, the position of the line FX is left undetermined. Hence the following porism.

A triangle being given in position, a point in it may be found, such, that any straight line whatever being drawn through that point, the perpendiculars drawn to this straight line from the two angles of the triangle, which are on one side of it, will be together equal to the perpendicular that is drawn to the same line from the angle on the other side of it.

6. TRIANGLES and parallelograms of unequal altitudes are to each other in the ratio compounded of the ratios of their bases and altitudes.

7. If ACB, ADB be two triangles upon the same base AB, and between the same parallels, and if through the point in which two of the sides (or two of the sides produced) intersect two straight lines be drawn parallel to the other two sides so as to meet the base AB (or AB produced) in points E and F. Prove that AE= BF.

8. In the base AC of a triangle ABC take any point D ; bisect AD, DC, AB, BC, in E, F, G, H respectively : shew that EG is equal to HF.

9. Construct an isosceles triangle equal to a given scalene triangle and having an equal vertical angle with it.

10. If, in similar triangles, from any two equal angles to the opposite sides, two straight lines be drawn making equal angles with the homologous sides, these straight lines will have the same ratio as the sides on which they fall, and will also divide those sides propor. tionally.

11. Any three lines being drawn making equal angles with the three sides of any triangle towards the same parts, and meeting one another, will form a triangle similar to the original triangle.

12. BD, CD are perpendicular to the sides AB, AC of a triangle ABC, and CE is drawn perpendicular to AD, meeting AB in E: shew that the triangles ABC, ACE are similar.

13. In any triangle, if a perpendicular be let fall upon the base from the vertical angle, the base will be to the sum of the sides, as the difference of the sides to the difference or sum of the segments of the base made by the perpendicular, according as it falls within or without the triangle.

14. If triangles A EF, ABC have a common angle A, triangle ABC : triangle AEF:: AB.AC: AE. AF.

15. If one side of a triangle be produced, and the other shortened by equal quantities, the line joining the points of section will be divided by the base in the inverse ratio of the sides.

II. 16. Find two arithmetic means between two given straight lines. 17. To divide a given line in harmonical proportion.

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