Question

Timele David Tetteh: Math 43 Time Limit: 3:00:00 Time Left:2:16:01 The har Analyze the theorem and its proof. Match each missing justification in column on the with corresponding statement from the column on the right. The following theorem can be viewed as a special case of the inscribed angle theorem. The proof below does not use the inscribed angle theorem. If the vertices of AA BC lie on a circle and AB is a diameter of that circle, then ZACB is a right angle. N Proof: Since COAC Since AB is a diameter of the circle, the midpoint of AB must be the center of the circle. Denote the center by O. Then OA OB OC by (Justification 1). By the (Justification 2) we have ZOAC ZOCA and 20BC = LOCB. Leta =(LOAC) and B - (ZOBC). Then 2a + 2B = 180° by the Justification 3). Thus & + B = 90° and ZACB is a right angle. 5 1. by contruction 9 4 2. That's what we want to show O 30 tv RA AC 20 F3 F7 2 # 3 $ 4 % 5 6. & 7 8 Q W E R. Y

Answer 1

Ab must be the center B А 0 OA, of the cirde denoted the center by o OB and oc are the radious of the circle Then. Since AB is a diameter of the circle The mid point of Again vertices of a ABC are lie on the Circle is, The 4 vertices AB Clie on the Circle. Now Joining The points cand o Then A = B = OC Now & AOC is isosceles triangle as OA=0C Therefore LOAC = LOCA = a (say) and & Boc is also isosceles triangle as <OBC= LOCB=B (Soy) Now for the LOACA LOCAT LAOC = 180 and for the 4 boc LOBC & LOCB & LBOC - 180 adding these two equation <O AC + LOCA+ CAOC + <OBC+LOCBAC BOC = 360° OB = 0c Therefore 4 AAOC as COAC = LOCA=d. LOB(=LOCB=B er, at & + LAOC + BAB+ <Boc 360° or, 20+2B + LAOC & BOC = 360° It is clease that (AOC of LBOC= 180° so a, 24+2B + 180º = 360° a 2012 B = 180° ar diß - 90 Therefore CACB = 90° XTB Tas LACB = LOCAT OCB