Trial Examination II B C D E D. Chandler 13. and CD intersect at point K,AD CB. , andab. 1. Suppose AB. CD. Prove that i) AK. ii) DAB BCD iii) ADK CBK

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1 Trial xamination II handler Suppose and intersect at point K,, and. rove that i) K K K ii) iii) K K 2. In suppose that -- and. rove that >. 3. Suppose 1 > 4. rove that 2 is acute In suppose, bisects, bisects and intersects at. rove that bisects. 5. Suppose Int( ), is the perpendicular bisector of, is the perpendicular bisector of, and G is perpendicular to. rove that G bisects. G

2 Supplementary roblems hapter 10 handler In the Saccheri quadrilaterals and WXYZ assume X Y that WZ and WY. rove that XYZ. W Z 2. Suppose is a Saccheri quadrilateral, M is the midpoint of, and N is the midpoint of. rove that N MN and MN. M 3. orollary. ( i) In the Saccheri quadrilateral,. (ii) Saccheri quadrilateral is convex. 4. Suppose is Saccheri, --, --G, =, = G, =, and is right. rove G. G 5. Suppose and are congruent Saccheri quadrilaterals where --. rove that if,, and are collinear then is a rectangle. 6. an you do this (replicate a Saccheri quadrilateral vertically) in the absence of the parallel postulate? Justify your answer.

3 handler In the Saccheri quadrilateral let M be the midpoint of and let N be the midpoint of. On MN locate point so that N MN. Through let L be a line X Y L perpendicular N and on L locate points X and Y so that X = M and Y = M. Through X and Y draw perpendiculars XU and YV to L, intersecting in U, V. y U N V considering various diagonals prove that XUVY so that = U and = V. How do you explain this in view of problem 6? M 8. In suppose,, and are right angles. xtend to so that = and extend to so that =. rove that is Saccheri. 9. In suppose, are midpoints of,, respectively. Let, G, H be the feet of perpendiculars to from,,, respectively. Suppose -G-. a) rove that H is Saccheri. G H b) rove that H = Suppose is Saccheri with right angles at and. Let be the midpoint of. Through draw the perpendicular to, intersecting at. rove that is the midpoint of. 11. Suppose is Saccheri with right angles at and. Show that = if and only if m( ) = 90.

4 Supplementary roblems hapters 11, 12. handler Let,, be the midpoints of the sides of as shown. rove that (and thus similarly, and ) and = ½ (and similarly, = ½ and = ½). 2. With the hypotheses of #1 prove that. 3. In point on is such that bisects. rove that =. Hint: Through draw a line L parallel to and extend to intersect L at. omplete the proof by exploiting and the fact that a certain triangle is isosceles. 4. In suppose is right, is a median, and. Show that,, and = ½. 5. Suppose is a trapezoid with, /, and =. If and are the midpoints of and, respectively. rove that and 6. Show that there is no SSSS congruence theorem for quadrilaterals. What about SSSS? SS? SSS? Justify your answers.

5 7. retend that this is prior to hapter 11 (no arallel ostulate). What is wrong with the following argument? 1 2 handler 17 What is actually proved? Let be an arbitrary triangle and suppose --. Label the various angles as shown. Let x be the 3 4 sum of the angle measures in a triangle. Then In : x = m( ) + m( ) + m( ) = m( ) + m(1) + m(2) + m( ) In : = m( ) + m(1) + m(3) In = m(4) + m(2) + m( ). dding the last two equations gives 2x = m( ) + m(1) + m(2) + m(3) + m(4) + m( ). rom this subtract the second equation to get x = m(3) + m(4) = 180 since 3 and 4 form a linear pair.

6 Supplementary roblems hapter 13 handler median of a triangle is a segment from a vertex to the midpoint of the opposite side. a) rove that a median separates the interior of a triangle into two regions having equal area. b) In suppose that and are medians intersecting at. Use a) to show that a( ) = a( ) and a( ) = a( ). 2. The area of an ellipse having major and minor semi-axes of lengths a and b, respectively, is πab, generalizing the area of a circle. Would you expect that its circumference to be π(a+b), again generalizing the circumference of a circle? Hint: onsider an extreme situation. b a 3. uplicate resident James. Garfield s elegant proof of the ythagorean Theorem. a c b 4. uplicate the Indian mathematician haskara s ( ) proof of the same. c a b 5. uplicate Leonardo da Vinci s ( ) proof of the same. b c a Note: haskara s proof is reported to have consisted of the above (#4) sketch together with the word ehold!. In fact, both his and Garfield s proof require a little algebra to see. On the other hand, Leonardo s sketch, if looked at correctly, truly is a ehold! proof. Trial xamination III.

7 handler Suppose is a Saccheri quadrilateral with right angles at and. Let be the midpoint of and let be the foot of the perpendicular from to. rove that is the midpoint of. 2. L 1, L 2, L 3, and L 4 are lines lying in the same plane and that L 1 L 2, L 3 L 4, and L 2 L 3. Show that L 1 L 4. L 1 L 2 L 3 L 4 3. Let be an arbitrary convex quadrilateral. Let M, N, O, be the midpoints of,,,, respectively. rove that MNO is a parallelogram. M N O 4. rhombus is a parallelogram with adjacent sides congruent. rove that for the rhombus that: a) The diagonals and are perpendicular. b) area( ) = ½.

8 Supplementary roblems hapter 16. handler Let O be a point interior to the circle and let L and M be two distinct lines through O, L intersecting in and Q, M intersecting in R and S. Show that R O S O OQ = OR OS. Q 2. Let be a circle with center and suppose T is a point exterior to. Let and be points on with T tangent to. Then m( T) = ½ arc(). T 3. Let the line L intersect the circle at points and Q. Let T be a point on L exterior to and let be a point of with T tangent to. rove that T 2 = T TQ. Q T 4. rove: convex quadrilateral can be inscribed in a circle (concyclic) if and only if its opposite angles are supplementary. Hint: Supplementary roblem 8 from hapters 6-7 may prove useful. 5. Two vertices of a triangle and the feet of the corresponding altitudes are concyclic. 6. Suppose is concyclic and its diagonals and intersect at. rove that.

9 Supplementary roblems hapter 19 handler Let be a circle with center (an S-point) and radius r (a surd). escribe briefly how to construct a circle with center whose area is twice that of. 2. If Moise had defined a surd angle, statement (1) in the middle of page 292 could read (?) There is a surd angle with measure 20. (?) How would you define surd angle? 3. etermine whether each of the following is true or false. riefly justify your answers. a) line with a non-surd slope could contain two S-points. b) If L is an S-line, L could contain a point having one surd coordinate and one non-surd coordinate. c) 6 8 is a surd. d) Let be an S-circle with center (0,0). There exists a surd t such that the points (t,t), (t,-t), (-t,t), (-t,-t) all lie on. e) If x is not a surd then neither is x 2. f) If X 3 is a surd then so is x. g) Given an S-line L and a non-s-point, we can construct (with compass and straight-edge) a line through parallel to L. h) S-circles can contain points having exactly one surd coordinate. i) 3 3 is a surd. j) It is possible with compass and straight-edge to subdivide a circle into 9 congruent subarcs. 4. If θ = 360/7 then cos 3θ = cos 4θ = 8 cos 4 θ 8 cos 2 θ + 1. Use this to show that 16 cos 4 θ 8 cos 3 θ 16 cos 2 θ + 6 cos θ + 2 = 0. Next, show 2 cos θ is a root of (x 2)(x 3 + x 2 2x 1), proving that θ is not constructible.

10 Some Special Theorems handler 22 Theorem. (oncurrence of ngle isectors) The angle bisectors of are concurrent (i.e., have a common point). roof. Let the angle bisectors of and intersect at and from R draw and, Q, R perpendicular to the sides of as shown. Then Q R and Q by H so that R (HL) and we see that is the bisector of. Q Note: Since, Q, R are perpendicular to the sides of and are equal in length (by T), is the center of a circle which is tangent to the sides of at the points, Q, R. This is the circle inscribed in. is usually called the incenter of. orollary 1. If is the incenter of then m( ) = 90 + ½m( ). roof m( ) = 180 (m( ) + m( )) = 180 ½(m( ) + m( )) = 180 ½(180 m( )) = 90 + ½m( ). orollary 2. If is on the angle bisector of in and m( Α) = 90 + ½m( ) then is the incenter of. roof. Let ' be the incenter of. y orollary 1 m( ') = 90 + ½m( ) = m( ) and by definition ' lies on the bisector of ' so that, ', and are collinear. If -'- then by roblem 8 on page 12 of the Supplementary Notes ' < while if --' then < '. oth cases contradict m( ') = m( ) so we must have = '.

11 handler 23 Theorem (rank Morley, 1899). The points of intersection, Q, R of adjacent trisectors of the angles of are the vertices of an equilateral triangle. roof. or convenience we define three numbers: α = 60 m( )/3 β = 60 m( )/3 γ = 60 m( )/3. Then α, β, γ are positive and α + β + γ = 120. R Q We will start with an arbitrary equilateral triangle XYZ and build up a triangle which is similar to and positioned to XYZ the same way that is positioned to QR. Let X' be the point on the opposite side of YZ from X such that m( YZX') = α = m( ZYX'). hoose Y' and Z' in a similar fashion relative to XZ andxy, respectively. Z γ α 60 γ β X' Y β α 60 γ β Z' Y' β α 60 γ X α Now m( ZYZ') + m( YZY') = γ β > γ + α β = 180, the uclidean arallel ostulate implies the rays Y'Z and Z'Y meet at a point. Likewise X'Z and Z'X meet at a point and Y'X and X'Y meet at a point. We will show that. Now m( ZX') = 180 (α β) = γ, and similarly, m( YX') = β, etc., establishing all angle measures as shown in the diagram above. urthermore, m( YZ) = 180 (2α + β + γ) = α = 60 α and similarly, m( XZ) = 60 β and m( XY) = 60 γ. Now ZX ' YX ' by the converse of the Isosceles Triangle Theorem, XZ XY, and XX ' XX ' so that XZX' XYX' by SSS. We conclude that XX ' bisects ZX'Y = X'. urthermore, m( X) = 180 a = 90 + (90 a) = 90 + ½(180 2a) = 90 + ½(m X') so (by the second corollary to the oncurrence of ngle isectors Theorem) X is the incenter of X' and we may conclude that XX' X.

12 In an exactly analogous way Z is the incenter of Z' so that X' X'X. Hence Z and X are handler 24 trisectors of. We conclude that m( ) = 3m( ZX) = 3(60 β) = m( ). Once again, in an exactly analogous manner we can show that m( ) = m( ) and m( ) = m( ). Thus. urthermore, Z R, X, and Y Q (all by ). y the definition of triangle similarity we have X = and, since and R Z, = = R Z. Therefore, XZ R by the SS similarity theorem. nalogously (once again), ZY RQ and YX Q. pplying all this similarity we get Q R R R = = = = = so that Q R. INLLY, analogously,r QR so that XY X X Z ZX XY QR is equilateral.

13 Trial inal xamination. handler Suppose is a Saccheri quadrilateral. rove that the segments joining midpoints of opposite sides are perpendicular: HG. You may use without proof that HG and HG. H G 2. Suppose S is a circle with center. If is a diameter H and H, H, and are all tangents then prove that area( H) = ½ H. I 3. Use the methods of to prove that it is possible to trisect a 90 angle. 4. Let have a right angle at and suppose at point with --. Show that = Suppose in that m( ) = 90, bisects where --, and. rove that and. If = 3 and = 6 what is?