Question

o is the point of concurrence of the internal angle bisector of a triangle ABC which meet its circumcircle at X Y and Z.then (Multi answer type)

a)Angle xyz are 90+a/2 ,90+b/2 and 90 + c/2
b)Angle xyz are 90-a/2,90-b/2 and 90-c/2
c)Angles boc,coa and aob are 90+a/2 ,90+b/2 and 90 + c/2
d)Angles boc,coa and aob are 90-a/2,90-b/2 and 90-c/2

Answer 1

Step-by-step explanation:

In △DEF,

∠D=∠EDF

But ∠EDF=∠EDA+∠FDA ....angle addition property

Now, ∠EDA=∠EBA and ∠FDA=∠FCA .....Angles inscribed in the same arc

∴∠EDF=∠EBA+∠FCA

=

2

1

∠B+

2

1

∠C

[Since BE is bisector of ∠B and CF is bisector ∠C]

∴∠D=

2

∠B+∠C

....(1)

Similarly, ∠E=

2

∠C+∠A

and ∠F=

2

∠A+∠B

...(2)

Now, ∠A+∠B+∠C=180

o

....angle sum property of triangle

∴∠B+∠C=180

∘

−∠A ...(3)

Similarly, ∠C+∠A=180

o

−∠B and ∠A+∠B=180

o

−∠C ...(4)

Substituting eq (3) in eq (1), we get

∴∠D=

2

180

o

−∠A

∠D=90

o

−

2

1

∠A

Similarly, ∠E=90

o

−

2

1

∠B and ∠F=90

o

−

2

1

∠C