University:
Felician University
Course:
MATH-231 | Geometry I
Academic year:

2022
Views:

371

Pages:

11
Author:

Clint W.

Notes Of Triangle Theorems Connecting the midpoints of two sides of а triangle creates а segment called а midsegment of the triangle. в Point Х is the midpoint of АВ. Point Yis the 1nidpoint of ВС. So, ХУ is а 1nidsegment of ЛАВС. There is а special relationship between а midsegment and the side of а triangle that is not connected to the midsegment. в Тriangle Midsegment Iheorem • The midsegment is parallel to the third side ofthe tтiangle. • The length oftl1e Inidsegment is halfthe length ofthe third side. ХУ 1 АС and ХУ= ½лс. Connecting each pair of 1nidpoints, you can ее that а triangle has three midsegments� ХУ, У2, and ZX are all n1idsegme11ts of ЛАВС. . Because Zis th e midpoint of АС, ХУ= AZ = ZC = ½лс. в Find х. 3х с:: -½Ls4) \ 2::> )< -:: ½ (34-) х -::::: 2 7 (х == 1 i:= 11 j / х+9 'z. 3х. � f l�-\-2-o') �Х.-::: 2 'i-- ...\- "' '2-У-== �Lx+cr) lO 4-)( L!:�o J = х+ 1 - - 6")( ::: <=\ l )( � D 1s the mldpolnt of АВ. Е 1s the mldpolnt of СВ. в . If mLA = 70" find mLBDE. +О , IfmLBED = 73, findmLC. �3 , If DE = 23, find АС.; 2 (z.3) :: 4(о 1 If с= 83, find DE. i L Ж3 ::: '-t l/::> А с PERPENDICULAR BISECTOR / CIRCUMCENTER Perpendicular Bisectors There are two u eful theorem to remember аЬонt perpendicнlar Ьisectors. Pe1-pendi cular вt Х is on tl1e perp endicular ector 1heorem blsector, so it is If а point is on the perpendicular Ыsector of а segn1ent, then it is equidistant from the endpoints of the segment. з ,,' , ,; х .. � equid.istant from the endpoints А and В. ,;' А-"-----В Converse ofthe Perpendicular Bisecto1· Iheo,�em If a point is equidistant from the endpoints of а seginent, tl1en it i оп the р erpendicнlar Ьisector of tl1e segment. Ftnd the alue of х. Because Х is equidistant fro1n the endpoints С and D, it is on the р erpendicular blsector of the segment. PERPENDICULAR BISECTOR / CIRCUMCENTER Тhе Circumcenter of а Triangle Sym agram rem Th р rp ndi нlar Ьi 1de t at tant fr s rp ndi ula Ы ctor Р ·, JJY 1d PZ t t А int th в с z What are the coordinates of the circumcenter of а triangle with the following vertices: А(-4, б) В(4, З) С(-2, -2) А(2, 7) В(10, 7) С(10, З) f\ ) 1/ � -r е I i.l 1 "'-.. / / ' � 11 "' "' f\ \'� .......... - ......_I �\ w / L-" Г', ',1" • / ,,r""" ,,,,,,,,,,,, \ I ........... _......,,... 'L\ � 7 r--..... ,,,/ ..t ANGLE BISECTOR / INCENTER Angle Bisectors There are two useful tl1eorems to remember about angle Ьisedors. Xis on the angle Ьisector and is therefore equidistant from the sides ofthe angle. Angle Bisector Тheorem If а point is on the Ы ector of an angle, then the point is eqшd:istant from the ides of tl1e angle. В ecause Х is in the intelior of the angle and js equidistant from the sides, Х is on tl1e angle Ьisector. Converse о{ the Angl Bisector IJ1eorern If ар oint in the interior of an angle is equidistant from the sides of an angle, then the point Е is on the angle Ьisector. Flnd the alu of х. 3х )(-=: 1,.... 4{10) qo (4х)О == 12._ 4 .J '-fx -== 2--x-f zь (2х + 2 Zx_-::::20 l Х-=10 l � 0}0 х10)+ 2C -r �-1° "2,o ANGLE BISECTOR / INCENTER Тhе lncenter of а Triangle 1heorem SymЪols Diogrom в The Ы ecto ol the angl of triangl з n urr nt ( t а point equidi tant frum the ide of tЬе triangle. AngJ Ы tor АР, ВР, and i: Р fl con ir nt ot Р. . = PZ !р r= Р А с z )'ои u1Ш pro1� 711 orcm 28 11 r ro.s 1: в Flnd the value of G с z А = • А с 7,,.,-х - 9 'Ь_-== Zx + 7Zx.-=- f- X:=-1: <= 3. s{ �< )( в -=- Х -t 4 -== 1( L-- J �+ 4-� 15 ? rz�-""::J--::: l ьD ::::� MEDIANS / CENTROID The Median А median of а triangle is а segment that runs from one vertex of the triangle to the midpoint of the opposite side. The point of concurrency of the medians is called the centroid. The median of ЛАВ aJe АМ, СХ, and BL. The centroid is point D. А The medians of а triangle intersect as а point two thirds of the distance from а vertex to the opposite side. This is the Concurrency of Medians Theorem. In дVУХ, the centrold lsZ. U е th dlagram to sol е the proЫems. lf XR = 24, find XZ and ZR V з<-24) = 5 Х,Ъ::: 2{�) -= l lt, 8 � \'2.-, .:::- =-- -___, lf VZ = 14, ftnd VP and ZP z ( t4)c: + lf XZ = 44, find XR and ZR {44 :=. l- 2.. f -------;z. ?:;/2. === z хе..� зс:22,) -= lo<,p lf ZO = 10, find YZ and УО 't � Z-(tь) -- 2о УСУ= 3(\о) -= �о MEDIANS / CENTROID Pis the centroid of ДМ О. МР = 14х + Ву. Write expressions to represent PR and MR. -\'(2,, ,::: -+ -у,_ -+ 4 м � --::. :::>L7 'i. + 4u) -- -· CS' с=: 7--t г<- --\: \� Fis the centroid of .6.АСЕ. AD = l5x2 + Зу. Write expressions to represent АР and FD. U е ·coordinate geometry to prove the follo\l ing statement. ------ \1о'х2 +� 3 - Дr;; :::: 2- (_S'л2- с+ '?s) /D r---1 .-.:= '6y._---t� --- - � 'Z- +� )Ох 2- -\- 'Z. ) J ALTITUDES / ORTHOCENTER The Altitude An altitude of а triangle is а segment that runs from one vertex perpendicular to the line that contains the opposite side. The orthocenter is the point of concurrency for the alti tudes. An altitude may Ье inside ог outside of the triangle, ог on а side of the triangle. ALTITU DES / ORTHOCENTER Determlne whether АВ ls а medtan, an altttude, or neither. 1. 0----- А ALTITIJD L3. А �7 2. А т 4. А '-- --- --- -------