Question

prove the following
1)

3.7) Corollary (Parallel CT). Let l, and l be coplanar lines and I a transversal. a. (Property C) 4 | l, if and only if a pair of interior angles on the same side of t are supplementary b. (Property T) Ift 1 l and 41 || 12, then t 1 l.

2)

3)

4)A parallelogram is a square iff it's diagonals are perpendicular and congruent.

5) the median of a trapeziod is parallel to each base

Answer 1

& You w a Ana Given a, & lo ane coplanar Linea and 'X' is transversali from the figure, L3 and 15, LA and 16 gode paines of interion angles on Same side of transvest sa 1. bet d, and lę are parallel - (4=27 cconresponding angles ) Li +(3 = 180' 1 also cl +CQ=180° Ý . Csale We have lat(4 = 1800 Since (4=27 . → Cat27 =180' 23+27= 180° (!' C&13) (23+15=180°7 ("(7:15) So, pair of interior angles on same side are supplementary if two Linea are parallel. Convesisly, 100% first take inted1007 angles pains in Supplementary: ie; (4+16+ 1800 be lonow Le+24=180° .: 28=267

- alipo 66+67= 180° 124=17) Similarly, cr=15, L3=18. A pair of corresponding angleze ale trual, if too liners are parallel. (6) Given , ttle and l, llda 1. We know, L2=90° di & de age parallel, ž ice 1. transversal and L1, L2 form - pair of conresponding angled on Same side, So caual. : LIECQ = 90° s t ime perpendiculaoy to deny & cas we Let SABC with right angle a have to prove CA and LB are complementary since ce is right angle. ie; 20=90' So from triangle sum theorem cat <b+LC=180- o La +66 +90 = 180 - Cat 2b = 90 tence ca & cb are complementary...

Let ABCD be a suadsilateral 1. bet we join AC clearly citl2 = CA O and (3+24 = cc from o ABC, we have LQ + 24 +28 = 180 > and from o Ape, we have Ci+20+ L3 = 180 → from 0 e 0 ZI+LQ+1 3+1 4+ LB +Los 360 ZA + LB ++CD=360 *-: 21728 LA 13+1+-te] flence sum of measure { of angles of any convex quadoilateral in 360° 1 o there is no rectangle frist . We know that, The Summit angleg of a Sacchez; quadrilateral are acute". we prove this by contradiction. Let floco a gectangle. Since ABCD is a Sacchers cuadorgla foral. ement above statement By The The Summit Summit an. angles are acute.

3.1 Theoremi This contradicts that the Summit angles a ale right angle as per defination of rectangle : Thus, No rectangle cript. 1 -3) - If possible set d, 4 da le; e, is not parallel to la Now, given d, tm and to Im Consider e,= AB * ср m = Ef and the following diagram. Since, t, Im then ZAEF = (Befogo. A M B and seem then LeFE - LEFO=90' c o . CAEF - (BEF - LOFE - LEFD=90 Therefore, d, ll da which contradicts and assumption d, it is 286, fill de CPoroved), lum

19 9Suppase that some pastallelog.nam pe seuale, - and then draw each diagonal, as shown below- Then, o len, O ACDN SBDC by . A SAT theoren (side-angle-side). This Cheorem applica because L AB A BD ~ Dc e c a by definition c of a Sauare. Square with diagonales Also, CA CD a LBDC since Both are right angles, again by I de fruition of a seuare. Having shown that o Aco a DB DC. We can then infor by cp CF C congruent parts of congruent figuree , that Ão e Be , which is to say that the diagonals are congruent. It only remains to eshow that the use diagonale goje peopendiculas). Since m2 c = 20 then m LCAD +mZCOA = 90 and Since BACD is isosceles ,

then m LCAD m LCAD-m LODA - 45 he Beme argument shook that mlocBr45 • Consider triangle OCDE and we see that m ZCED = 90 aq desioned. alow, to prove the other direction, faust assume that AD a BC and AD L BC . Since we are still operating under the hypothesis that are whole figure is a parallelogram, then we ponon that opposite sideęs acte Congruent and parallel, due to the corollary of the theorem foot parallebgram a . Theice AB & co. Thus, we can Ishow that o ADB » DEBD by sss ( side-side -side). the other two sidegs ade known to be congruent because we have assumed . AD A Be and 30 a 60 by reflexivity. From this we can infen LDAB LBCD by cpof. Noo alogo from the fact that . inter

and figure is a parallelogram with ABIED transversal ge, ule can infer from the z property. LOCB LCBA. This, together with the previous result, : implies LLBA » COAB. If we consider OEAB We see that 1 m CDAB+ m LCBA = 90 and from what llle have just seen, m 2 DAB= 45 = m LCBA. this argument can be used again, this time showing a CBA 2 ODAB to show that every small angle is 45 degrees and. There foge every interior angle of the paor alle logoram ię 90 degrees. All that remains, then, is to show that Some two adjacent edges of the rectangle are congruent : 1. But this follows immediately from the fact that A CBD iç isosceles with .LOCB a CDBC, due to the isosceles Triangle heorem. I

1. Consider the trapezoid ABCD with bases AB, CO and F, M & F 9.9e midpoints of AD, ÂCE CB Porove that EF DAB , FFOOD -- --- --- - -- -- - - AB I CO consider the traingles DADC , DABC The Line Joining the midpoints of two sides of a traingle is to the third side EMA CO and MF D AB Transitive property of congruence EM O ME Parallel postulate through fixed pointe theore is only one line parallel to the I given Two line. lines . O median are identical, EM = E MED AB = EFT AB EM O co EF I CD of a trapezoid is 11 to each bases The Sell

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