Therefore if through B, you draw BD to BA making the angle ABD to the given angle, join AD, bisect BD in E, join EA; it is evident that the line drawn thro' H parallel to BE is the line required to be drawn. For the angles AGF,ABD, are, also the angles GFA,BDA, .. GFA,GAF are,.. GF is GA, and since BD is bisected by AE, GF is also bisected by it, ... HG is half of GA, and the angle AGF is to the given angle,

PROP. 113, THEOR.

to

If the points of bisection (A,B,C,D) of any quadrilateral figure (such as EFGHI) be connected, the figure thus formed is a parallelogram.

Fig. 68.

Join EH, FI.

Then because in the triangle FEH, the sides FE and FH are bisected, AB is parallel to EH, also in the triangle IEH, since IE,IH are bisected, DC is parallel to HE, (prop. 7 of deduc)... DC is parallel to BA, for the same reason AC is parallel to BD, .', ABCD is a parallelogram.

PROP. 114, THEOR.

In figure sixty-nine, if the right lines [DG, EO, LF] be drawn, the triangles [DAG, EBO, LCF] thus formed, are equal to one another.

Produce AB to K; then the angle CBK is evidently= to OBE, but CBK is the supplement of CBA, .. OBE is the supplement of CBA. Then in the triangles BOE, BCA, the sides OB,BE are respectively to CB,BA, and the contained angles ABC,OBE supplemental, the triangles OBE,CBA are, (prop. 44 of deduc.) for the same reason the triaugles DAG and ABC are =, and ABC and FCL are, (by prop. 4, 1 Elr.)

PROP. 115, PROB.

Given the excess of the perpendicular of an equilateral triangle above half the side to construct it.

Fig. 70.

=

Upon the given excess AB describe an equilateral triangle ACB, draw CH at right angles to it, make HI=to HA and IE to IC, upon CE describe an equilateral triangle DEC and produce CA to F : CF will be at right angles to DE, and AF will be≈ to half CE; join AI and FI. Then because AH is to IH and AHI a right angle, each of the angles HAI,HIA is an angle of 45, .. the angle IAF is an angle of 75, and because the engle FCE is an angle of 30 and DCE an angle of 60, DCF is an angle of 30, .. CF is at right anles to DE and FE is half DE, ... FI is to EF, . FIE is an equilateral triangle, and since FIE is an angle of 60 and HIA of 45, FIA is an angle of 75 and is.. to FAL .. FA is to FI and .. to IE, ... CDE is the required triangle.

=

PROP. 116, THEOR.

If the hypothenuse [AB] of a right angled triangle [ABC] be produced, and perpendiculars [DM,OK] be let full on the produced line from the angles [FDA, LOB] of the squares described on the sides, those perpendiculars shall be respectively equal to their adjacent segments [AP,BP] of the hypothenuse, and the intercepts [MA,KB] between the perpendiculars and hypothenuse, are each equal to the perpendicular [CP] from the right angle on the hypothenuse.

Fig. 69.

=

Because AL is parallel to BO the angle OBK is to LAK and.. to BCP; then in the triangles BCP,OBK the angles OBK,BKO of the one are respectively to BCP,CPB, of the other and CB is to BO,... OK is to BP and KB to CP, (prop 26, 1 Elr.) for the same reason DM is to AP and MA to CP.

=

PROP. 117, THEOR.

In figure sixty-nine, the squares of the jorng lines [DG, EO and L] are equal to six squares of the hypothenuse [AB.]

Produce DA and OB to meet perpendiculars in S and N.

2

Because CBN is a right angle and also EBA, if NBÁ be taken away, the remainders EBN,CBA are =,... the triangles BNE,BCA are similar and equal, for the same reason ASG,ACB are similar and ≈, .. ON is bisected in B and also DS is bisected in A,.. (props. 43 and 47, 1 Elr.) the of OE is to four times the of OB, or BC the square of NE or CA, for the same reason the of DG is to four times the 2 of DA or of AC+ the ' of SG or of CB,.. the □rs of DG and OE together are to five times the rs of AC and CB together or to five times the of AB,.. the squares of DG, OE, FL together are to six times the of AB. FL is evidently to AB, (prop. 8, 1 Elr.)

2

PROP. 118, THEOR.

2

If two right angled triangles (ABC,ECD) be similar, the square described upon the sum of the hypothenuses is equal to the squares described upon the sum of the bases, together with the squares described upon the sum of the other two sides.

Fig. 71.

Place the given triangles so that the bases BC,CD may be in directum, produce BA 'till the produced part AL is to CE, join LE.

Since EC is and parallel to LA, LE is 7 and parallel to AC, .. AE is a parallelogram, .. the angles LAC,LEC are, also the angles BAC,CED are =, (by hypoth.) but LAC and BAC are to two right angles,.. LEC,DEC are to two right angles, .. LE, ED are in directum.

Then since LBD is a right angled triangle, the ' of LD is to the 2 J'rs of LB,BD, but LD is = the sum of the hypothenuses, BD the sum of the bases and LB the sum of the sides .. &c. &c.

PROP. 119. PROB.

In the greater side (BD) of a scelene right angled triangle (BDE) to find a point from which if two right lines be drawn, one to the opposite angle (BED) and the other to the hypothenuse (BE), their sum shall be a minimum.

Suppose the sum of EF,FG to be a minimum: produce ED 'ill DC is to DE, join FC, it is evidently = to FE; then since FC is to FE and the sum of FE,FG the least that can be drawn from any point in DB, the sum of CF,FG is the least that can be drawn from the point C to BE, .. CF,FG are in directum and at right angles to BE..

If then, you produce ED 'till DC is to it, draw CG at right angles to BE, join FE, the sum of FG, FE is less than the sum of any other two lines drawn from any other point in BD.

For, draw any other lines OL,OE, join OC,LC. Then CG is evidently to FE,FG, for the same reason CO,OL are to EO,OL, but CO,OL are greater than CL, and CL is greater than CG, because CGL is a right angle, CO,OL or EO,OL are much greater than CG or EF,FG, ... &c. &c.

PROP, 120. PROB.

In a given quadrilateral figure (DECG) to find a point from which the sum of the lines drawn to the angles

shall be a minimum,

Fig. 52.

- Draw the diagonals CD,GE: their sum is less than the sum of any other lines drawn from any point ✪ to the angles.

For DC is less than DO,OC together, . SG, SC, SE, SD, are together less than OD, OG, OC and OE together, ..the point S in which the diagonals cut one another, is the point required to be found.

PROP. 121, THEOR.

In figure sixty-nine, the right lines (DB,CG) cut one another at right angles.

ANALYSIS.

Suppose that Bs t, one of the angles at the point of section, is a right angle, it is to the angle t AG, and the angles Bts, G tA are = (prop. 15, 1 Elr).. the triangles G A t, B8 t are simile, if this is proved the proposition is true.

Then in the triangles BDA,GAC the sides BA and AD are respectively to GA and AC, and the angle GAC to BAD, .. the triangles are in every way, .. the angles AGC, ABD are =. Then in the triangles Bts, G At the angles & Bt, A G t are = (by demon.) and the angles At G, Bts are (by 15, 1 Elr) . the angles t s B, t A G are = (cor. 32, 1 Elr.) but t A G is a right angle, .. ts B is a right angle and .. the other angles at the point s are right angles.

PROP. 122. PROB.

To inscribe in a given square (DCG) a right line parallel to a right line (LG) given in position and equal to [FI] given in magnitude, not greater than the diameter of the square.

Fig. 11.

Draw from any angle BDE of the given DK parallel to LG, make DO to FL, through O draw OQ parallel to DE and through Q draw QP parallel to KD: it is to FL and parallel to FG; the demonstration is evident.

PROP. 123. THEOR.

If the sides of a square (ADFC) be bisected and right lines (a D, C c, d F, Ab) be drawn from the points of bisection to the opposite angles (FDA, CFD, DAC, ACF) the figure (eg ki) formed by those drawn lines is a square, and is the fifth part of the entire figure.