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Demonstration. Because ABC is an angle in a semicircle,
1. ABC is a right angle (III. 31).
3. BA is to AC as AD is to AF (VI. 4);
Now the angle ADF is also a right angle (hyp. and the angle BAC is either the same with DAF, or vertical to it; therefore
2. The triangles ABC, ADF are equiangular, and therefore
4. The rectangle BA, AF, contained by the extremes, is equal to the rectangle AC, AD, contained by the means (VI. 16).
Wherefore, if from one extremity, &c. Q.E.D.
If the angles at the base of a triangle be bisected by two lines that meet, and the exterior angles at the base, formed by producing the two sides, be similarly bisected, the two points of concourse and the vertex shall be in one straight line, which shall bisect the vertical angle.
Let ABC be a triangle, of which the angles at the base are bisected by the lines AD, BD, which meet in D, and let the exterior angles be bisected by AG and BG.
Then the two points D, G, and the vertex C, are in one straight line, which bisects the angle at C.
Construction. Draw DE, DF, DL, and GM, GK, GN perpendicular to the sides and sides produced.
Demonstration. Because in the triangles ADE, ADL, the angles at A are equal, and the angles at E and L are likewise equal, being right angles, and AD is common to them; therefore
1. AL is equal to AE, and DL to DE (I. 26).
may be proved in a similar manner, that
2. BL is equal to BF, and DL to DF;
3. AM is equal to AK, and GM to GK;
4. BN is equal to BK, and GN to GK. And since DE and DF are each equal to DL; 5. DE is equal to DF, and GM to GN.
Again, since in the triangles CED, CFD, the two sides CD, DE are respectively equal to CD, DF; and the angles at E and F are right angles; consequently the triangles are equal in every respect; and therefore
6. The angle at C is bisected by CD.
It may be proved in a similar manner, that the triangles CMG, CNG, are equal in every respect, and therefore
7. The angle at C is bisected by CG ;
8. The line CD coincides with the line CG.
GEOMETRICAL EXERCISES ON BOOK I.
If two opposite sides of a parallelogram be bisected, and two lines be drawn from the points of bisection to the opposite angles, these two lines shall trisect the diagonal.
Let ABCD be a parallelogram of which the diagonal is AC. Let AB be bisected in E, and DC in F; also let DE, FB be joined, cutting the diagonal in G, H.
Then AC is trisected in the points G, H.
Construction. Through E draw EK parallel to AC, and meeting FB in K (I. 31).
Demonstration. Because EB is the half of AB, and DF the half of DC; therefore
1. EB is equal to DF;
and these equal and parallel straight lines are joined towards the same parts by DE and FB; therefore
2. DE and FB are equal and parallel (I. 33). And because AEB meets the parallels EK, AC, therefore
3. The exterior angle BEK is equal to the interior angle EAG (I. 29).
For a similar reason,
4. The angle EBK is equal to the angle AEG (I. 29). Hence in the triangles AEG, EBK, there are the two angles GAE, AEG in the one, equal to the two angles KEB, EBK in the other, and one side adjacent to the equal angles in each triangle-namely, AE equal to EB; therefore
5. AG is equal to EK (I. 26);
but EK is equal to GH (I. 34); therefore 6. AG is equal to GH.
By a similar process, it may be shown that
7. GH is equal to HC.
Hence AG, GH, HC are equal to one another, and therefore 8. AC is trisected in the points G, H.
Wherefore, if two opposite sides, &c. Q.E.D.
2. On a given straight line as base, to describe an isosceles triangle having each of its equal sides double of the base.
3. A line drawn bisecting the angle contained by the two equal sides of an isosceles triangle, bisects the third side at right angles.
4. On a given straight line to describe an isosceles triangle, of which the perpendicular height is equal to the base.
5. If two triangles on opposite sides of the same base, have the two sides equal which are terminated in one extremity of the base, and likewise those equal which are terminated in the other extremity of the base, the two triangles are equal in all respects. To be proved without using I. 8.
6. Given two points, one on each side of a given straight line; find a point in the line such that the angle contained by two lines drawn to the given points may be bisected by the given line.
7. The opposite angles of a rhombus are equal.
8. If two circles cut each other, the straight line which joins their centres bisects that which joins the points of intersection.
9. From every point of a given straight line, the straight lines drawn to each of two given points on opposite sides of the line are equal prove that the line joining the given points will cut the given line at right angles.
10. From a given point to draw a straight line to a given straight line, that shall be bisected by another given straight line.