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From B draw BD parallel to CP, and BE parallel to CQ to meet AC at D and E.
Then because CP is parallel to BD,
therefore the angle CBD is equal to the angle BCP, and the angle CDB to the angle ACP;
but the angle BCP is equal to the angle ACP, therefore the angle CBD is equal to the angle CDB, and therefore CD is equal to CB.
Again, because PC is parallel to BD,
therefore AP: PB :: AC: CD:
but CD is equal to CB,
therefore AP: PB:: AC: CB.
In like manner it may be shewn that
AQ: QB:: AC: CB,
IV. 9, Cor. 2.
Conversely, let the base AB of the triangle ABC be divided internally at P and externally at Q
in the ratio AC: CB.
then shall CP and CQ be the bisectors of the interior and exterior vertical angles at C.
For the bisectors have been shewn to divide AB internally and externally in the ratio AC: CB,
and there is only one point in which AB can be divided internally, and only one point in which it can be divided externally in the ratio AC: CB,
and P, Q are these points of division,
therefore, by the Rule of Identity, CP bisects the vertical angle ACB, and CQ the exterior vertical angle BCD.
Ex. 19. If two triangles have one angle of the one equal to one angle of the other, and a second angle of the one supplementary to a second angle of the other, then the sides about the third angles are proportional.
ABC is a triangle which has its base BC bisected in D. DE, DF bisect the angles ADC, ADB, meeting AC, AB Prove that EF is parallel to BC.
in E, F.
*Ex. 21. OA, OB are two tangents to a circle whose centre is C; OPQ any straight line through O cutting the circle in P and Q, and AB in R. If N be the intersection of AB and OC, prove that NR bisects the angle PNQ, and consequently that PQ is divided harmonically at O and R.
*Ex. 22. O is a fixed point, OPQ any straight line through O, cutting a given circle in P and Q; R is taken on PQ such that OQ: OP :: RQ: PR. Shew that R is on a fixed straight line.
*23. If three straight lines drawn from a point cut wo parallel straight lines in A,B,C and A',B',C' respectively, prove that AB BC : : A'B' : B'C'.
#24. If two parallel straight lines AB, A'B' are divided both internally, or both externally, in the same ratio at C and C' respectively, prove that AA', BB', CC' meet in a point.
*25. The straight line DEF meets the sides BC, CA, AB at the points D,E,F respectively, and is equally inclined to AB and AC. Prove that BD: CD :: BF: CE.
26. D is a point in the side AC of a triangle ABC, E is a point in AB. If BD,CE divide each other into parts in the ratio 4 1, then D,E divide CA, BA in the ratio 3 1.
27. If AD bisect the angle BAC and meet BC in D, and DE, DF bisect the angles ADB, ADC, and meet AB, AC in
E, F respectively, prove that the triangle BEF is to the triangle CEF as BA is to AC.
28. If two isosceles triangles have equal vertical angles, their altitudes are to one another as their bases.
*29. Parallelograms about the diagonal of a parallelogram are similar to the whole, and to one another.
30. Similar and similarly placed parallelograms, which have a common angle, have their diagonals in the same straight line. 31. AEKH, KFCG are parallelograms about the diagonal AC of a parallelogram. Prove that EF, AC, GH are either parallel or meet in a point.
32. If ABC be a triangle inscribed in a circle, and the tangent AD at A meet BC produced in D, the diameters of the circles about ABD, ACD are as AD to CD.
33. Two circles touch internally at O. A straight line touches the inner circle at C, and meets the outer in A,B; and OA,OB meet the inner circle in P,Q. Shew that
OP OQ AC: CB.
34. A series of triangles is constructed such that the straight lines drawn from the vertices to the middle points of the opposite sides of each are equal to the sides of the next. Prove that the alternate triangles of the series are similar.
35. One of the angles B of a right-angled isosceles triangle ABC is trisected by lines which meet the straight line AMN drawn from A the right angle perpendicular to the base BC in M and N, and CN produced cuts AB in E. Shew that EM is parallel to BN.
36. If three similar rectilineal figures be so placed as to have their corresponding sides parallel, their three centres of similarity, when taken two and two, lie in a straight line.
THEOR. 10. If four straight lines are proportional the rectangle contained by the extremes is equal to the rectangle contained by the means;
and, conversely, if the rectangle contained by the extremes is equal to the rectangle contained by the means the four straight lines are proportional.
Let OA, OB, OC, OD be four straight lines, such that OA: OB:: OC: OD:
then shall the rectangle contained by OA and OD be equal to the rectangle contained by OB and OC.
Let OA and OB be placed along one of the arms of a right angle, OC, OD along the other arm;
complete the rectangles AD, BC,
and let BD be the common part of AD, BC.
Because rectangles of equal altitude are to one another as their bases,
therefore the rectangle AD is to the rectangle BD as OA is to OB. For the same reason, the rectangle BC is to the rectangle BD as OC is to OD;
but OA: OB:: OC: OD,
therefore the rectangle AD is to the rectangle BD as the rectangle BC is to the rectangle BD,
therefore the rectangle AD is equal to the rectangle BC, that is, the rectangle contained by the extremes OA, OD is equal to the rectangle contained by the means OB, OC.
Conversely let the rectangle contained by OA and OD be
equal to the rectangle contained by OB and OC:
then shall OA be to OB as OC is to OD.
Let the same construction be made as before.
Then, because the rectangle AD is equal to the rectangle
Hyp. therefore the rectangle AD is to the rectangle BD as the rectangle BC is to the rectangle BD.
But the rectangle AD is to the rectangle BD as the base OA is o the base OB.
And the rectangle BC is to the rectangle BD as the base OC is to the base OD,
therefore OA : OB :: OC : OD.
COR. If three straight lines are proportional, the rectangle contained by the extremes is equal to the square on the mean;