EXERCISES.

1. Shew that every quadrilateral is divided by its diagonals into four triangles proportional to each other.

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

3. From a 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.

5. If the bisector of the angle B of a triangle ABC meets AD at right angles, shew that a line through D parallel to BC will bisect AC.

6. From B and C, the extremities of the base of a triangle ABC, 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 that 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.

PROPOSITION 3. THEOREM.

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 have, the straight line drawn from the vertex to the point of section shall bisect the vertical angle.

E

X

ABC let the BAC be bisected by AX, which

In the meets the base at X;

Through draw CE par to XA, to meet BA produced

Again, because XA is par1 to CE, a side of the ▲ BCE,

that is,

.. BX XC :: BA : AE;

BX: XC :: BA : AC.

VI. 2.

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

For, with the same construction as before,

because XA is par' to CE, a side of the ▲ BCE,
.. BX: XC :: BA: AE.

But by hypothesis BX XC :: BA : AC;

.. BA: AE :: BA : AC;
.. AE = AC;

.. the ACE the AEC.
But because XA is par1 to CE,

.. theXAC the alt. ACE.

BAX = the int. opp.

and the ext.

VI. 2.

v. 1.

I. 5.

I. 29.

AEC;

I. 29.

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 be 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 B and 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 this proposition 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.

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 bisector and the extremities of the base shall have to one another the same ratio as the remaining sides of the triangle have:

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

F

X

In the ABC let BA be produced to F, and let the exterior CAF be bisected by AX which meets the base produced at X:

Again, because XA is par1 to CE, a side of the ▲ BCE,

.. BX: XC :: BA : AE;

that is, BX XC :: BA : AC.

Constr.

VI. 2.

Conversely, let BX: XC :: BA: AC, and let AX be joined: then shall the FAX = the XAC.

For, with the same construction as before, because AX is par' to CE, a side of the

.. BX XC :: BA: AE.

But by hypothesis BX: XC :: BA : AC;

.. BA: AE :: BA: AC;

BCE,

VI. 2.

v. 1.

1. 5.

Propositions 3 and A may be both included in one enunciation as follows:

If the interior or exterior vertical angle of a triangle be bisected by a straight line which also cuts the base, the base shall be divided internally or externally into segments which have the same ratio as the sides of the triangle:

Conversely, if the base be divided internally or externally into segments which have the same ratio as the sides of the triangle, the straight line drawn from the point of division to the vertex will bisect the interior or exterior vertical angle.

EXERCISES.

1. In the circumference of a circle of which AB is a diameter, a point P is taken; straight lines PC, PD are drawn equally inclined to AP and on opposite sides of it, meeting AB in C and D ;

2. From a point A straight lines are drawn making the angles BAC, CAD, DAE, each equal to half a right angle, and they are cut by a straight line BCDE, which makes BAE an isosceles triangle: shew that BC or DE is a mean proportional between BE and CD.

3. By means of Propositions 3 and A, prove that the straight lines bisecting one angle of a triangle internally, and the other two externally, are concurrent.