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If a straight line is drawn parallel to one side of a triangle, it cuts the other sides, or those sides produced, proportionally.

Conversely, if the sides, or the sides produced, are cut proportionally, the straight line which joins the points of section, is parallel to the remaining side of the triangle.

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Let XY be drawn par to BC, one of the sides of the A ABC.

Then shall BX : XA :: CY : YA.

Join BY,

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I. 37.

СХ. Now the A' BXY, CXY are on the same base XY and between the same parls XY, BC ;

the A BXY = the A CXY ;

and AXY is another triangle ; ... the A BXY : the A AXY :: the A CXY : the A AXY. V. 4. But the ABXY : the A AXY :: BX : XA,

VI. 1. and the A CXY : the A AXY :: CY : YA;

... BX :XA :: CY : YA.

v. 1.

:

Conversely. Let BX :XA :: CY: YA, and let XY be joined.

Then shall XY be pard to BC.

As before, join BY, CX.
By hypothesis, BX : XA :: CY : YA;

but BX : XA :: the A BXY : the A AXY, VI. 1.

and CY: YA :: the A CXY : the A AXY ; ... the A BXY : the A AXY :: the A CXY : the A AXY. v. 1. ... the A BXY = the A CXY;

V. 6. and these triangles are on the same base and on the same side of it; .. XY is par to BC.

I. 39. Q.E.D.

EXERCISES.

1. Shew that every quadrilateral is divided by its diagonals into four triangles whose areas are proportional3.

2. If any two straight lines are cut by three parallel straight lines, they are cut proportionally.

3. From the point E in the common base of two triangles ACB, ADB, straight lines are drawn parallel to AC, AD, meeting BC, BD at F, G: shew that FG is parallel to CD.

4. In a triangle ABC the straight line DEF meets the sides BC, CA, AB at the points D, E, F respectively, and it makes equal angles with AB and AC: prove that

BD : CD :: BF : CE.

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5. In a triangle ABC, AD is drawn perpendicular to BD, the bisector of the angle at B: shew that a straight line through D parallel to BC will bisect AC.

6. From B and C, the extremities of the base of a triangle ABC, straight lines BE, CF are drawn to the opposite sides so as to intersect on the median from A: shew that EF is parallel to BC.

7. From P, a given point in the side AB of a triangle ABC, draw a straight line to AC produced, so tha it will be bisected by BC.

8. Find a point within a triangle such that, if straight lines be drawn from it to the three angular points, the triangle will be divided into three equal triangles.

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If the vertical angle of a triangle be bisected by a straight line which cuts the base, the segments of the base shall have to one another the same ratio as the remaining sides of the triangle.

Conversely, if the base be divided so that its segments have to one another the same ratio as the remaining sides of the triangle, the straight line drawn from the vertex to the point of section shall bisect the vertical angle.

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In the A ABC, let the 2 BAC be bisected by AX, which meets the base at X.

Then shall BX : XC :: BA : AC.

Through c draw CE par' to XA, to meet BA produced at E.

1. 31. Then because XA and CE are par, .:. the 2 BAX = the int. opp. L AEC,

I. 29. and the L XAC = the alt. - ACE.

I. 29. But the 4 BAX = the _ XAC;

Нур. ... the L AEC = the 2 ACE; .. AC=AE.

I. 6.

Again, because XA is par to CE, a side of the A BCE,

BX : XC :: BA : AE ; that is,

BX: XC :: BA : AC.

VI. 2.

Conversely. Let BX :XC :: BA : AC; and let AX be joined.

Then shall the L BAX = the < XAC.
For, with the same construction as before,
because XA is par to CE, a side of the A BCE,
BX : XC :: BA : AE.

VI. 2. But, by hypothesis, BX : XC :: BA : AC; .. BA : AE :: BA : AC;

V. 1. .. AE = AC; ... the L ACE = the L AEC.

I. 5. But because XA is par to CE, ... the L XAC = the alt. _ ACE.

I. 29. and the ext. 2 BAX = the int. opp. L AEC; 1. 29. .. the L BAX = the L XAC.

Q.E.D.

EXERCISES.

1. The side BC of a triangle ABC is bisected at D, and the angles ADB, ADC are bisected by the straight lines DE, DF, meeting AB, AC at E, F respectively: shew that EF is parallel to BC.

2. Apply Proposition 3 to trisect a given finite straight line.

3. If the line bisecting the vertical angle of a triangle is divided into parts which are to one another as the base to the sum of the sides, the point of division is the centre of the inscribed circle.

4. ABCD is a quadrilateral: shew that if the bisectors of the angles A and C meet in the diagonal BD, the bisectors of the angles Band D will meet on AC.

5. Construct a triangle having given the base, the vertical angle, and the ratio of the remaining sides.

6. Employ Proposition 3 to shew that the bisectors of the angles of a triangle are concurrent.

7. AB is a diameter of a circle, CD is a chord at right angles to it, and E any point in CD; AE and BE are drawn and produced to cut the circle in F and G: shew that the quadrilateral CFDG has any two of its adjacent sides in the same ratio as the remaining two.

a

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If one side of a triangle be produced, and the exterior angle so formed be bisected by a straight line which cuts the base produced, the segments between the point of section and the extremities of the base shall have to one another the same ratio as the remaining sides of the triangle.

Conversely, if the segments of the base produced have to one another the same ratio as the remaining sides of the triangle, the straight line drawn from the vertex to the point of section shall bisect the exterior vertical angle.

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In the A ABC let BA be produced to F, and let the exterior – CAF be bisected by AX which meets the base produced at X. Then shall BX: XC :: BA : AC. Through c draw CE par' to XA,

I. 31. and let CE meet BA at E. Then because AX and CE are par, ::. the ext. L FAX = the int. opp. L AEC, and the L XAC = the alt. – ACE.

I. 29. But the L FAX= the L XAC;

Hyp. ... the L AEC=the ACE;

... AC=AE. Again, because XA is par to CE, a side of the A BCE,

Constr. .. BX : XC :: BA : AE ;

VI. 2. BX; XC :: BA ; AC,

I. 6.

that is,

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