Geometric Proof

�_t_'{�----- Name ____ Date _______ Class _______ Practice А Geometric Proof Write the letter of the correct justification next to each step. (Use one justification twice.) Given: HJ is the Ьisector of LIHK and L1 1. 2. 3. 4. 5. = L3. А. Definition of L Ьisector HJ is the Ьisector of LIHК. в. Given L2 L1 с. Transitive Prop. of L1 L3+ � L2=L3 --proof, each step in the proof is on the left and ln а 11лJD - �(1\u м the reason for the step is on the right. = В 8 = n Fill in the Ыanks with the justifications and steps listed to complete the two-column proof. Use this list to complete the proof. L1 :::L2 Def. of straight L L1 and L2 are straight angles. 6. Given: L1 апd L2 are straight angles. Prove: L1 L2 Proof: = 1. а. L I v,nG{ L2 Statements g�t- arf 5.-+rct. \ 2. mL1 = 180 °, mL2 = 180 ° 3. mL1 = mL2 4. с. .1 C�fЧ1k� 1. Reasons Given 2. Ь. d.e Р.. о { �-i а. ; а hi L '5. 3. Subst. Prop. of = 4. Def. of = .& L/ ·;:L2- Follow the plan to fill in the Ьlanks in the two-column proof. 7. Given: L1 and L2 form а linear pair, and 1/ L3 and L4 form а linear pair. Prove: mL1 + mL2 + mL3 + mL4 = 360 ° Plan: The Linear Pair Theorem shows that L1 and L2 are supplementary and LЗ and L4 are supplementary. The definition of supplementary says that mL1 + mL2 = 180 ° and mLЗ + mL4 = 180 ° . Use the Addition Property of Equality to make the conclusion. /'\ Statements 1. L1 and L2 form а linear pair, and L3 and L4 form а linear pair. 2. L1 and L2 are supplementary, and L3 and L4 are supplementary. 0 3. с. d) L: 1 iмL1=tfl.) 1'У\ { -� + т L Ч 1 .fO = 4. mL1 + mL2 + mL3 + mL4 = 360 ° 1. а. 2. ь. Reasons g ivt:t"--L1V\..itн 3. Def. of supp. .& 4. d. А 1,{1.� р (,1. � у' � 1'У\ -- nmo· с� Kf '/ Name ______________ Date ______ Class ______ Practice В Geometric Proof Write а justification for each step. Given: АВ = EF, В is the midpoint of АС, and Е is the midpoint of DF . .. ' . F � А В 1. В is the midpoint of АС, and Е is the midpoint of DF . 2. АВ = ВС , and DE = EF. rf-e ( i 1 f (У) 1/f рб, 1'\ t d; Г. t, ·( ( о·" 9 (t/f ,'+ 3. АВ = ВС, and ОЕ = EF. Sf 4. АВ + ВС = АС, and ОЕ + EF = DF. L 5 . add. po<:.t • h S l-1 /Pi- л h� PD L) , 11-е У'-, ( р, L nл re ) 5. 2АВ = АС, and 2EF = DF. 6. АВ=EF 7. 2АВ = 2EF 8. AC=DF 9. АС= DF Fill in the Ыanks to complete the two-column proof. 10. Given: LHKJ is а straight angle. К! Ьisects LHKJ. Prove: LIKJ is а right angle. Proof: Reasons Statements 1. а. L 1--1 t' ::r 1 \. {,/ 2. mLHKJ = 180 ° 3.с. �-t{(\ t(1h. t Ct f\t11 е.. �""1 Ь ,�е l 1 ':> LH tl- Т 4. LIKJ = LIKH 1. Given 2. ь. (,�о { о} � WO,l С1 \л 1 / 3. Given 4. Def. of L Ьisector =л 5. mLIKJ = mLIKH 5. Def. of 6. d. mL 1:l(T � М Lirr\ :- mLH к.т 6. L Add. Post. 7. 2mLIKJ = 180 ° 7. е. Subst. (Steps ? 1S-1 8. mLIKJ = 90 ° 8. Div. Ргор. of = 9. LIKJ is а right angle. 9. f. Gle f · (,' f ь r •�\\ t. ) L t 'IL ,;:; , 1 I _______ Name ___..:....\ )---=L,;_� Date ______ Class ______ Practice С Geometric Proof Write а two-column proof. 1. Given: The sum of the angle measures in а triangle is 180 ° . Prove: mL1 = mLЗ + mL4 (9 (j) .S-\tfu�м) m j__ '2- � (() L �1 m L L/ = 1 КОс. L 1 (., f\t( L 2- а� r(.L.L / , n.R и./ ptt 1 L2- с. re.. L � {f) @ I С-(\(,/ rY\ L 1 -1 fYI L 2: �L� I fD i pp . ,/\ � ,� J v(<;.1'�cat-DY\ Q) g \ "-€ 1\ (jJ С\\ vf- "'j (j) @ . ti L I Y\R O t' p f ( --1h W\ dR_ /. t { {\.f)P 1. L-fи_ h @:.>uf)St, /3 /_3 1 S С\ r, 1htL @dJ'f cr v ,91-\t @ r, 6) LJ L3 1 /-;. 1 -� et r-t?. 1/ е r h 1�i:,_( {.. , _s (i д 6) С)\{ f ,_ cJJ ( . 11е. v tч ьr- JL J /_, Name _____...... K t Y ___. '-'-------- Date ______ Class ______ •i• ProЫem Solving Geometric Proof 1. Refer to the diagram of the stained-glass window and use the given plan to write а two-column proof. Given: L1 and LЗ аге supplementary. L2 and L4 аге supplementary. L3:::L4 Prove: L1 Plan: (D L1 1 j_ "2 1 = L2 Use the definition of supplementary angles to write the given information in terms of angle measures. Then use the Substitution Property of Equality and the Sublraction Property of Equality to conclude that L1 L2. = Ъ с.. с--е s.l,f р ' . L L/ Qrf Svff 1. L m '-- 3 ...: 1 ои о rnL 1 -t- vf""\(') mL l 1 mL\ t мLЧ м == /fQ vY1LЧ L 3 _:. Г{'\L?. t (j;) S i, � Ь st. @1t�� - , P Ot L 3 '"\,, L-Ч � U-.J df f' о{� { L L мL 3 - rYl t1h�t.Po� S {j;) f L L m 2" . r vn L \ � rn L Ч = \'Y'\L (5) S1,,{ blr. рга р 01 и YY\LI = rnL2, g ,r\, ® d_p { c!J ro t;�'. The position of а sprinter at the starting Ыocks is shown in the diagram. Which statement сап Ье proved using the given information? Choose the best answer. 2. Given: L1 and L4 аге right angles. А LЗ = L5 С mL1 + mL4 = 90 ° D mLЗ + mL5 = 180 ° � 3. Given: L2 and LЗ аге supplementary. L2 and L5 аге supplementary. � G L2 = L5 Н LЗ and L5 аге complementary. J L1 and L2 аге supplementary. {\_f,_�-------- Date ______ Class ______ Name ___ Challenge Prove lt! ln а proof, you can often determine the Given information from the figure. Write the information that is given in each figure. Then make а conjecture about what you could prove using the given information. Е 2. 1. А L1 ,ve �'\ ', в .L-1 � L ч LA &: (), cve : L 2 = L3 ,'scl у; �h{ ( �n�v-11:() (Y\(,t � \., CJ'� L б , ve V'- � f f -= [Т F б -=: Т µ - р, ov-f ) •. л.., Е Н ::: Е L::1 3. Write а two-column proof. Given: LKLM and LNML аге right angles. L.2:L.3 L.1 = L.4 Prove: L КLЦ LUµ L L KLH l\..,L�HL т L KL Н = т L ,\Jf'v\L I rn L KLM =- (Y)L.I -1 rлL L L-°:) 1 мLЧ (Y'I = L И iJ L rn 2 L� -----M u.; 9 ·,\Je /;\ @ ,...__ ftC)"' f L --п._УУ). (J ck ( . о .{ · G) й.nc.Jli..7 C(c.t"t/. "'--' poS't •