Saved Bookmarks
| 1. |
Ex 10.2 Q13 |
|
Answer» Thanks In quadrilateral ABCD circumscribes a circle with centre O. Join OP,OQ, OR and OS.In tri. OPB and tri OQB. Angle OPB= Angle OQB = 90PB =QB (Theorem 10.2)OB=OB (common )Tri.OPB=tri.OQB ( by RHS congruency rule)ang.POB=ang.BOQ (1=2)(by CPCT)similarly ang.QOC=ang.COR (3=4) ang.ROD= ang.DOS (5=6) ang.SOA=ang.AOP (7=8)therefore , Sum of angles of a quadrilateral is 360.angle 1+2+3+4+5+6+7+8=360 2 (ang.2+3)+2 (6+7)=360 (because ang.1=2,3=4,5=6,7=8) and similarly, 2 (ang.1+8)+2 (4+5)=360(2+3)+(6+7)=180 and (1+8)+(4+5)=180ang.BOC+ ang.AOD=180 and ang.AOB+ang.COD=180 H.P. |
|