AD is the bisector of angle BAC meets BC in O . arrange AD, BD, CD in descending order .
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Picture of the triangle is not given. So I draw one. Hope that is right.
see diagram.
Let E be the midpoint of BC. ∠CAB is a right angle. so ∠DAB = 45° = x = ∠DAC.
We start with a triangle with ∠B > ∠C. Hence, ∠B > 45° as they are complementary angles.
Clearly CD > EC and EC > BD.
We also see that EC = EA (as this is the circumradius).
EA > AD. so AD < EC.
Hence, CD , AD, BD is the descending order.
see diagram.
Let E be the midpoint of BC. ∠CAB is a right angle. so ∠DAB = 45° = x = ∠DAC.
We start with a triangle with ∠B > ∠C. Hence, ∠B > 45° as they are complementary angles.
Clearly CD > EC and EC > BD.
We also see that EC = EA (as this is the circumradius).
EA > AD. so AD < EC.
Hence, CD , AD, BD is the descending order.
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kvnmurty:
clik on thanks..
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