14. Prove the following construction for inscribing a square in a triangle ABC. On the side BC remote from A describe a square BCDE. Join AD, AE meeting BC in F, G. The square on FG is the square required.

15. A common tangent to two circles cuts the join of the centres externally or internally in the ratio of the radii.

16. Two circles intersect in A and B. Any chord CBD cuts the circles in C and D. Prove that the ratio AC: AD is constant for all

positions of the chord.

17. ABC is a triangle right-angled at A; a point D is taken in the hypotenuse BC such that CB : BA = BA : BD. Prove that AD is perpendicular to BC.

18. The line AB being divided at the points C and D, so that AB: AC AC: AD and any line AE (equal to AC) being drawn through A, prove that EC bisects L BED.

19. ABC is a straight line, and BD, CE are any two parallel lines on the same side of ABC. Points D and E are taken such that BD: CE = AB AC; prove that ADE is a straight line.

20. Show how to divide an arc of a circle into two parts whose chords are in a given ratio.

21. From a given point on the circumference of a circle draw two chords which are in a given ratio and contain a given angle.

22. ABC is a triangle which is not obtuse-angled. If the perpendicular AD from A to BC is a mean proportional between BD and DC, prove that BAC is a right angle.

23. ABC is a triangle which is not obtuse-angled, and AD is drawn from A perpendicular to BC. If CA is a mean proportional between CD and CB, prove that BA is a mean proportional between BD and BC.

24. Find the locus of points at which two given circles subtend equal angles.

25. Find a point at which three given circles subtend the same angle.

26. ABCD is a parallelogram, L, M and P are points in AD, CD, BD respectively, LP and MP meet BC, BA respectively in No and 0. Prove that LM is parallel to ON.

27. ABCD is a quadrilateral. On the side of AB remote from C, LBAE is made equal to CAD, LABEL ADC, and EC is joined. Prove that LECAL BDA.

28. ABCD is a parallelogram, and P and Q are any two points such that PQ is parallel to AB, PA and QB meet at R, and PD and QC meet at S; show that RS is parallel to AD.

300. THEOREM [RT.3]. In a right-angled triangle, if a perpendicular be drawn from the right angle to the hypotenuse, the triangles on each side of the perpendicular are similar to the whole triangle and to each other.

In ▲ ABC (Fig. 433):

Given that L BAC

=

R, and that AD 1 BC,

Required to prove that As ABC, DBA, DAC are similar.

T.1.

Thus As ABC, DBA are equiangular.

In As ABC, DAC,

=

Thus As ABC, DAC are equiangular. i. e. As ABC, DBA, DAC are equiangular. Hence As ABC, DBA, DAC are similar.

COROLLARIES. (i) DB: DA DA: DC,

=

i. e. DA is the mean proportional between DB and DC.

i. e. BA is the mean proportional between BD and BC.

i. e. CA is the mean proportional between CD and CB.

S.1.

S.1.

S.1.

S.1.

[These three results are easy to remember if it is noted that the three lines mentioned in Corollary (i) have a common extremity at D, the three mentioned in Corollary (ii) have a common extremity at B, and those in Corollary (iii) have a common extremity at C.]

NOTF. These three corollaries are equivalent to the theorems already proved in the corollary to Theorem RT.2 (§ 154), and Ex. IXXXIX, 13. The same reference symbol [RT. 3] has accordingly been used for all these theorems.

301. THEOREM [P.7].-If two parallel straight lines are cut by a series of transversals which all pass through one point, then the intercepts formed on the two parallel straight lines are proportional.

302. THEOREM [L.10]. The locus of a point within a given angle whose distances from the arms of the angle are in a constant ratio is a straight line through the vertex of the angle.

Given that PM: PN is a constant ratio k (Fig. 436),

Required to prove that the locus of P is a straight line through 0.

Cons. Let p be any one position of the moving point, so that pm: pnk. Take any other point P on Op, or Op produced. Let pm, PM, pm, PN be the perp. distances, from OA, OB.

Thus any point on Op satisfies the required condition. Hence Op is the required locus.

COROLLARY. -If PM = k. PN +7 where k and are constants, then the locus of P is still a straight line.

Let FG be a fixed line drawn || OA at a distance 7 (Fig. 437); also let FG meet PM at L. Then LM = 1.

Thus if PM k. PN + 1, PL = PM - 1=k. PN.

Hence the locus of P is a straight line through F.

303. THEOREM [Pn.5]. Two similar polygons can always be divided into similar triangles.

In Fig. 438, ABCDE . . ., abcde are similar polygons whose ratio of similitude is k, and in which the vertices A, B, C,. correspond respectively to the

vertices a, b, c,

...

In the same way ▲ ADE ||| ▲ ade, and so on.