Let the vertex of an angle ABC be located outside a circle and let the

1.	Let the vertex of an angle ABC be located outside a circle and let the sides of the angleIntersect equal chord AD and CE with the circle Prove that ABC is equal to half theDifference of the angles subtended by the cnords AC and DE at the centre
Answer» In triangle AOD and COE,OA=OCOD=OEAD=CEhence,they r congr.tnow,OAD=OCE (CPCT)ODA=OEC ( " )also,OAD=ODAi.e. OAD=OCEO=DA=OEClet them be 'x'now,in tri.OACOA=OC,so,it's an iso.tri.,let their equal ang's be 'a'lllylet the equal ang's of ODB be 'y'ADEC is a cyclic quad.now,AOC=180-2aDOE=180-2yDOE-AOC=180-2y-(180-2a)= -2y+a / 2a-2y2a-2y= 2a-(360-4x-2a) =2a-360+4x+2a 4a+4x-360ang.B=180-(180-x-a)-(180-x-a) =180-180+x+a-180+x+a=2x+2a-180i.e,measure of ang.B/difference = 2x+2a-180/4x+4a-360 =1/2 thanks👍👍👍👌👌👌👌

Let the vertex of an angle ABC be located outside a circle and let the sides of the angleIntersect equal chord AD and CE with the circle Prove that ABC is equal to half theDifference of the angles subtended by the cnords AC and DE at the centre

Answer»

In triangle AOD and COE,OA=OCOD=OEAD=CEhence,they r congr.tnow,OAD=OCE (CPCT)ODA=OEC ( " )also,OAD=ODAi.e. OAD=OCEO=DA=OEClet them be 'x'now,in tri.OACOA=OC,so,it's an iso.tri.,let their equal ang's be 'a'lllylet the equal ang's of ODB be 'y'ADEC is a cyclic quad.now,AOC=180-2aDOE=180-2yDOE-AOC=180-2y-(180-2a)= -2y+a / 2a-2y2a-2y= 2a-(360-4x-2a) =2a-360+4x+2a 4a+4x-360ang.B=180-(180-x-a)-(180-x-a) =180-180+x+a-180+x+a=2x+2a-180i.e,measure of ang.B/difference = 2x+2a-180/4x+4a-360 =1/2

thanks👍👍👍👌👌👌👌

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