Ch-circle
Q-ABCandADC are two triangle with common hypotenuse AC.prove that angel CAD =angelCBD
Q-prove that a cyclic parallelogram is a rectangle.
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1) ang.CBD=ang.CAD ----------------theorem 10.9
2) let ABCD be a cyclic llgm
in a llgm diagonals bisect;
so---DO=BO and AO=CO-------------------- 1
it is only possible when the diagonals are the diameters of the circle
now considering triangles DAB , ABC ,BCD and CDA
they all lies in the semi circle .i.e. angles a,b,c and d are equal to 90.as they lie on semi circle
Now,its a llgm wuth all angles of 90 digree and hence a rect.
2) let ABCD be a cyclic llgm
in a llgm diagonals bisect;
so---DO=BO and AO=CO-------------------- 1
it is only possible when the diagonals are the diameters of the circle
now considering triangles DAB , ABC ,BCD and CDA
they all lies in the semi circle .i.e. angles a,b,c and d are equal to 90.as they lie on semi circle
Now,its a llgm wuth all angles of 90 digree and hence a rect.
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