Pythagoras' Theorern and Its ApplicationsTheorern 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{lL
Trang 1Pythagoras' 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;
_,
Trang 2a2 +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
Trang 3Exanrple 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,
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,
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.
Trang 4Solution 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
Trang 5Exarnple 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
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
Trang 66.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.
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