TOÁN TIẾNG ANH

Pythagoras' Theorern and Its Applications Theorern L (Pythagoras' Theorem) For a right-angled triangle with tw,o legs a,h antl lrypotenuse c, the sum of squares of legs is equal to the s{lLtare of its /ty- prstennse, i.e. a2 ,*b2 - c2. Theorem ll. (Irwerse Theorem) If the lengths a,b,,c of three sides of a trinngle hqve the relation a2 + b2 : c2, then the tiangle must be a right-angled triangle with tyvo legs a,b and hypoterutse c. When investigating a right-angled triangle (or shortly, right triangle), the fol- lowing conclusions are often used: Theorenr III" A triangle is a right triangle, if and onty if the median on one side is lzalf of the side. Theorenr IY. If a right triangle has an interior angle of size 30", then its opposite Ieg is half of the htpotenuse. Example Example L. Given that the perimeter of a right angled triangle ls (z + rA) the meclian on the hypotenuse is 1 cm, find the area of the triangle. soiution The TheoremIII implies that AD - BD : cD - 1, so y'B [-et AC : b, BC - a, then c111; _, a2 +b2 :22 :4 and a+b -\/6. Therefore 6 - (a +b)' : a2 +b2 +2ab,so ab_T _t, the area of the triangle ABC it ;. Example 2. As shown in the figure, lC : BD :2.5 cm. Find AC. therefore, by Pythagoras' Theorem, " PCz : 9 : CQ, + PQz, ICQP: 9OO. Hence IAPB - ICQB - 90o + 45o - 135o \ i 90o, lL\,,: 72, CD - 1.5 cm, Solution From D introdu ce D E -L AB,intersecti ng AB at E . When we fold up the plane that ACAD lies along the line AD,thenC coincides with E, so AC - AE, DE : CD - 1.5 (cm) By applying Pythagoras' Theorem to ABED, BE JED, - DE, - \Ezb -'L25- z (.*). A Letting AC : AE - n cmand applying Sthagoras' Theorem to LABC leads the equation (r+2)':12+42', 4r-12, .'.r-3. Thus AC :3 cm. Example 3. As shown in the figure, ABC D is a square, P is an inner point sr:ch thatPA PB: PC:1:2:3. Find IAPB indegrees. Solution Without loss of generality, we assume that PA : 1,, PB : 2, PC - 3. Rotate the AAPB around B by 90o in clock- wise direction, such that P + Q,A + C, then AB PQ is an isosceles right triangle, therefore D Pe' - 2Pr,2 - 8,cQz - PA2 :1, P,/.' rQ a; I I I B Exanrple 4. (SSSMO(I)/2003) The diagram shows a hexagon ABCDEf, rnade Lip of five right-angled isosceles triangles ABO, BCO,CDO, DEO,EFO, and a triangle AOF, where O is the point of intersection of the lines BF and AE. Given that O A : B cm, find the area of A.AO F in cmz. Solution From oc _ #roB-(i)roA-toA, oE _ hoc-IoA-2(cm). Since RiAt.F'O - RIAABO, EF:oF-!o"-J-gn. 4 4\/z Let FG -L AE at G, then f'G - hOF - tOA: 1cm. Thus, the area of AAOF, SaAoF,is givenby steop -t oo.FG- 4 (cm2). Example 5. (Formula for median) In LABC, AM is the median on the side BC. Frove that AB2 + AC2 : 2(AM2 + n N127. Solution Suppose that AD -L BC at D. By Pythagoras' Theorem, A82 BD2+AD2-(BM+MD)z+AD2 BIUI2 +28X,[ .MD + MD2 + AM2 _ MD2 B]VTZ + AM2 +2BM . ]/ID, Similarly, we have AC2 : CIVT2 + A]VIz _zMC .]VID. B Thus, by adding the two equalitie,s up, since B M - C M , AB2 + ACz :2(AM' + BM'\. Note: When AM isextended to.E such that ABEC is a parallg.iogram, then the fornrula of median is the same as the parallelogram rule: AB2 + BEz + EC2 + C4' : AE2 + BC2. Exarnple 6. In the figure , lC - 90o , lA - 30o, D is the mid-point of AB and DE I AB, AE - 4 cm. Find BC. Solution Connect B,E. Since ED is the perpendicular bisector of AB, - 300 - 300, Now let BC _ r crn, then from Pythagoras' B Theorem, (2")':tr2+621a2-!2 +r-{r2-2tfrcm. CEA Thus, BC : ZtR cm. Exarnple 7. For aABC,o is an innerpoint, and D, E,F are on BC,CA,AB respectively, such that OD L BC, OE L CA, and OF L AB. prove that AF2 + BD2 + CE2 : BF2 + DC2 + AE2. Solution By applying the Pythagoras' Theorem to the triangles O AF, O B F , OBD, OC D,, OC E and OAE, it follows that A BE : AE, sa IEBD - IEBA - lA - 30o ,ICBE: 60o .'.CE : *na - DE - +AE -2 cm. AFz+BD2+CE2 - AO2 - OFz + BO2 - OD2 + CO2 - OE2 : (BQ' - OF,) + (CO2 - ODr) + (AO2 - OE ) - BF2 + DC2 + A82. T'he conclusioir is proven. Exarnple 8. In the diagram given below P is an interior point of aABC , p p1 _L AB, PPz )- BC, PP3 L AC, and BP1 - BPz, CP2 - Cps, prove thar AP7 - APs. Solution For the quadrilatenl AP1BP, since its two diagonals are perpen- dicular to each other, AP? + BPz _ AF2 * ptF2 + BF2 + pF2 _ AP2 + BP?. By considering AhC P and PC PzB respec- tively, it follows similarly that . AP2+Cp|-Ap|+pC2, BP? + PCz : pBz + Cpi. B Then adding up rhe rhree equalities yields Ap? - Ap|, Apt - Ap* Ps Exarnple 9. In square ABCD, M is the midpoint of AD and l/ is the midpoint of l\,'I D . Prove that IIY BC : 2lAB M . Solution LetAB - BC : CD - DA - a.LetEbethemidporntof CD. Let the lines AD and BE tntersect at F. By symmetry, we have DF - CB : a. Since A nght triangles AB M and C B E are syfilmetric in the line BD, IABIVI : ICBE. lt suffices to show INBE - IEBC, and for this we only need to show INBF - IBFIV since IDFE : IEBC. By assumption we have MN F D l _, I I 12. 'ln nx - Xo, .'. rrB - On ttre other hand, NF:!no*o-Xo, so l/F - BIU{,hence lI{BF - IBFN. rXa'+ o' - Xo. Testing Questions (A) r.{ (cirnl,ry1995) ln LABC, lA - 90o, AB - AC, D is a point on EC. Prove that BDz + C D2 - 2AD2 . 2. { Given that RTAABC has a perimeter of 30 cm and an area of 30 cm2. Find the lengths of its three sides. :1.( trn tlre RIAABC, lC: 90o, AD isthe angle bisector of lAwhich inter- sects BC atD. Given AB - 15 cm, AC:9cm, BD: DC :5:3. Find the distance of D from AB. 4. In the right triangle ABC, lC : g0o , BC : L2 cm, AC - 6 cm, the per- pendicular bisector of AB intersects AB and BC at D and -E respectively. FtndCE. *- /1 ,f u^ 5. In the rectangle 'ABCD, CE L DB a,t E, BE : 1U, and CE- 5 cm. Find the length of AC . -' 1 2, i '*' 4 6.F In AABC,IC: 90o, D is the mid-point of AC.Prove that AB2 +BBC2 - 48D2. In the right triangle ABC, lC :90o,, E, D arc points an AC and BC, respectively. Prove that AD2+BE2:AB2+D82. (CHNMOL/1990) LABC is an isosceles triangle with AB - AC - Z. Thereare 100points Pt,Pzl ,Proo ontheside BC.Wntemi - AP? + BPt,.PiC (i:1,2, ,100), findthevalue of m1 lm,z + *mrc0. In LABC, lC - 90o, D is the midpoint of AB, E, F are two points on AC andBC respectively, and DE L DF.kovethat EF2 : AE2+8F2. (CHINA/1996) Given that P is an rnner point of the equilateral triangle ABC , such that PA : 2, PB - 2tE, PC :4. Find the iengttr of the side of LABC. Testing Questions (B) (SSSMO(I)/2003/Q8) AB is a chord in a circle with center O and radius 52 cm. The point M dlides the chord AB such that AM - 63 cm and NIB - 33 cm. Find the length OM incm. (CHINA/I996) ABC D is a rectangle, P is an inner point of the rectangle such that PA:3, PB : 4, PC : 5, find PD. Determine whether such a right-angled triangle exists: each side is an integer and one leg is a multiple of the other leg of the right angle. (AHSME lLgg6) In rectangle ABC D, lC is trisecte dby C F and, C E,where closest to the area of the'rectangle ABC D? (A) 110, (B) 120, (C) 130, (D) 140, (E) 150. (Hungary 11912)Let ABC D bea convex quadrilateral. prove that AC L B D if and only if AB2 + CD2 : ADz + BC2. -{ (. 8.' I.t 10. 1. 2. J. 4. 5.