Math 3 Review Sheet Ch. 3 November 4, 2011 Review Sheet: Not all the problems need to be completed. However, you should look over all of them as they could be similar to test problems. Easy: 1, 3, 9, 10, 12, 13, 17 Medium: 2, 4, 5, 6, 11, 14, 18 Hard: 6, 8, 15, 16 Very Challenging: 7 Tests are made up of mostly medium problems with some easy problems and some hard problems. 1. What additional information is needed to prove these triangles congruent? What postulate would you then use to prove triangles congruent? E F A B C D Solution: If you had AE = CD then you could use SAS. If you had E = C then you could use AAS. If you had BE = CF then you could use Hypotenuse-Leg. Finally, if you had F = B you could use ASA.
2. Prove the following: (You may not use the Isosceles Triangle Theorem or its converse) Given: 1 = 2 A BC = BC Prove: ABC is Isosceles. B 1 2 C Solution: We know 1 = 2 and that BC = BC. Therefore we can prove ABC = ACB by ASA. Then by CPCTC, AB = AC. If a triangle has two congruent legs, it is isosceles; thus ABC is isosceles.
3. Prove the following: Given: ADC is isosceles A B is the midpoint of AD B C Prove: ABC = DBC D Solution: It is given that ADC is isosceles, so because isosceles triangles have congruent legs, AC = CD. We are also given that B is the midpoint of AB. Midpoints divide a segment into two congruent segments, so AB = BD. We now have three sets of congruent sides so by SSS ABC = DBC.
4. Prove: Given: L is trisected. L F LA is isosceles. Prove: LMF = LOA M F A O Solution: It is given that L is trisected, thus MLF = LF A = LAO because when an angle is trisected it is divide into three congruent parts. It is also given that F LA is isosceles, so because isosceles triangles have congruent legs and base angles, LF = LA and LF A = LAF. Then from the diagram we can say LF M$ LF A and LAO$ LAF. So we can say that LF M = LAO because if two angles are supplementary to congruent angles, then they are congruent. Therefore, by ASA, LMF = LOA.
5. Prove: Given: T is the midpoint of MS P Q P MT and QNT are right 1 = 2 R S MR = SN 1 2 Prove: P = Q M T N Solution: It is given that T is the midpoint of MN. So because a midpoint divides a segment into two congruent parts, MT = T N. We also know that 1 = 2 and that P MT and QNT are right. Becuase all right angles are congruent, P MT = QNT. Then RMT is complementary to 1 and SNT is complementary to 2 because they form right angles. So RMT = SNT because if two angles are complementary to congruent angles then they are congruent. It is also given that MR = SN which means we can prove RMT = SNT by SAS. Then by CPCTC ST N = RT M. So we can now show P MT = QNT by ASA. Therefore by CPCTC P = Q.
6. Prove: Given: CB = CD A B ABD = EDB C D CB and CD trisect ACE. Prove: ABE = EDC E Solution: It is given that CB = CD and since they are two legs of a triangle, the triangle is isosceles, so CBD is isosceles. Because CBD is isosceles and isosceles triangles have congruent base angles, CBD = DBC. Then it is given that ABD = EDB so by the subtraction property, if you subtract congruent angles into congruent angles, the resulting angles are congruent, ABC = CDE. Now CB and CD trisect ACE, and when an angle is trisected it is divided into three congruent angles, so ACB = BCD = DCE. In conclusion, by ASA we know that ABC = EDC.
7. Prove (this one is challenging) Given: AD = DB A B AE = BC F CD = ED E G H C Prove: AF B is isosceles. D Solution: We know AD = DB, AE = BC and CD = ED. Thus, AED = BCD by SSS. Then by CPCTC we can say EDA = CDB. Also, by the reflexive property ADB = ADB so by the addition property in which if you add two congruent angles to the same angle the resulting angles are congruent, ADC = BDE. Then by SAS ADC = BDE. Then by CPCTC DAC = EBD. Because AD = BD, ABD is isosceles because if a triangle has two congruent legs, it is isosceles. Also, if a triangle is isosceles, its base angles are congruent: DAC = DBA. Then by the subtraction property, which says if you subtract congruent angles from congruent angles, the resulting angles are congruent, we have BAF = ABF. Then by the converse of the isosceles triangle theorem, which states that if a triangle has congruent base angles then it is isosceles, we can conclude that BAF is isosceles.
8. Prove: Given: AD = CF B E BAC = DF E G ABC = DEF Prove: DGC isosceles A D C F Solution: It is given that BAC = DF E and that ABC = DEF. Then because AD = CF, we can say by the addition property (adding the same segment to congruent segments gives congruent segments) that AC = DF. Then by AAS ABC = F ED. So by CPCTC, GDC = GCD which makes DGC isosceles using the converse of the isosceles triangle theorem.
9. Prove: Given: F H bisects GF J and GHJ. G F 2 3 1 4 H Prove: F G = F J J Solution: It is given that F H bisects GF J and GHJ, so because a bisector divides an angle into two congruent parts we can say: 2 = 1 and 3 = 4. Then by the reflexive property F H = F H. So by ASA F GH = F JH. Then by CPCTC F G = F J.
10. Prove: Given: 5 = 6 JHG = O GH = MO Prove: J = P J 5 6 G H M P O Solution: We know that 5 = 6. From the diagram we know JGH$ 5 and P MO$ 6. Then because two angles that are supplementary to congruent angles are congruent, we can conclude that JGH = P MO. We are also given that JHG = P OM and GH = MO. So by ASA JGH = P MO. Then by CPCTC J = P.
11. Prove: Given: N is comp to NP O N S S is comp to SP R NP O = SP R N = SP Prove: NOP = SRP O P R Solution: It is given that NP O = SP R and that S is comp to SP R and N is comp to N P O. Thus by the complement theorem: if two angles are complementary to congruent angles then they are congruent, N = S. It is also true that N = SP. Therefore by ASA, NOP = SRP.
12. Prove: Given: 1 is comp to 3 2 is comp to 4 1 2 3 4 Prove: 1 = 4 Solution: From the diagram we can see 2 and 3 are vertical angles and because all vertical angles are congruent, 2 = 3. Then we are given 2 is comp to 4 and 1 is comp to 3. Therefore by the complement theorem: if two angles are complementary to congruent angles they are congruent, we can conclude that 1 = 4.
13. Prove: Given: P K and JM bisect each other at R. P M R Prove: P J = MK J K Solution: We know that P K and JM bisect each other at R. When a segment is bisected it divides a segment into two congruent parts, thus P R = RK and JR = RM. Also, from the diagram we can conclude that 3 and 4 are vertical angles, and because vertical angles are congruent, 3 = 4. So by SAS P RJ = KRM. Then by CPCTC P J = MK.
14. Prove: (Do not use the definition of isosceles or the isosceles triangle theorem) Given: JG is an altitude and a median F F J = HJ. G J Prove: F JG = HJG H Solution: It is given that F J is both a median and an altitude. Because F J is an altitude we can say that F J F H and perpendicular lines imply right angles so F GJ and JGH are right angles. Then JGH = F GJ, because all right angles are congruent. Because F J is a median, it meets F H at its midpoint G. Then F G = GH because a midpoint divides a segment into two congruent segments. And by the reflexive property JG = JG. Therefore by SAS we can conclude F JG = HJG.
15. Prove: Given: JK = MK J OP bisects JK and MK O Prove: JO = P K M P K Solution: It is given that JK = MK and that OP bisects JK and MK. When a segment is bisected it is divided in half, so 1JK = JO and 1 MK = MP. Therefore we know 2 2 JO = P K because the division property says like divisions of congruent segments are congruent.
16. Prove: Given: HO = MO H M JO = KO HJ is an altitude of HJK O MK is an altitude of MKJ Prove: MKJ = HJK J K Solution: It is given that MK is an altitude of MKJ and HJ is an altitude of HJK. Because altitudes are perpendicular to the side of the triangle they intersect, they form right angles with that side, so HJK and MKJ are right angles. This makes HJK and MKJ right triangles. It is also given that JO = KO and HO = MO. So by the addition property JM = HK. Also, by the reflexive property, JK = JK. Therefore HJK = MKJ by Hypotenuse-Leg. So by CPCTC, MJK = HJK.
17. Prove: Given: ADB = ADC A ABC is isosceles Prove: AD is a median B D C Solution: It is given that ADB = ADC and that ABC is isosceles. Then because isosceles triangles have congruent legs and congruent base angles, B = C and AB = AC. Thus ABD = ACD by AAS. Therefore by CPCTC, BD = DC. So if a point divides a segment into two congruent segments that point is a midpoint, so D is a midpoint. And then we can conclude that AD is a median because it is drawn from the angle of triangle to the midpoint of the opposite side.
18. Prove: Given: KG = GJ F G H 2 = 4 1 is comp to 2 3 is comp to 4 F GJ = HGK Prove: F G = HG K 1 2 4 3 J Solution: It is given that F GJ = HGK and we know by the reflexive property that KGJ = KGJ. Thus by the subtraction property F GK = HGJ. It is also given that 2 = 4 and 1 is comp to 2 and 3 is comp to 4. Then we can conclude that 1 = 3 because if two angles are complements of congruent angles then they are congruent. Then KG = GJ so by ASA GF K = GHJ. Then by CPCTC F G = HG.