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Question

Let the circles with centre P and Q touches each other at point A.Let the extended chord AB intersects the circle with centre P at point E and the chord BC touches the circle with centre P at the point D.
Then prove that the ray AD is the angle bisector of CAE.


Solution


Construction:- Join EDLet ADM=x and CAM=ySince, AM=DM (tangents from a external point are equal)So,  AMD is an isosceles triangleADM=DAM=x  (angles opposite to equal sides)Now, In ADCACB=CAD+ADC   (Exterior angle property)=DAM+CAM+ADC=x+y+x=y+2xCAM=ABC=y  (angles in the alternate segment)Now, In ABCABC+ACB+CAB=180°CAB=180°-y-y-2xCAB=180°-2x-2y        ...(i)DAC=DAM+MAC=x+y    .....(ii)Now, EAD+DAC+CAB=180°    (EAB is a straight line)EAD=180°-180°+2x+2y-x-y   (from(i) and (ii))EAD=x+yNow, EAD=DAC=x+ySo, AD is the bisector of CAE

Mathematics
Mathematics (Geometry)
Standard X

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