Question

Q.1.Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angles of the triangle DEF are 90° – (½)A, 90° – (½)B and 90° – (½)C.​

Answer 1

Step-by-step explanation:

In △DEF,

∠D=∠EDF

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

Now, ∠EDA=∠EBA and ∠FDA=∠FCA

∴∠EDF=∠EBA+∠FCA

= 1∠B+1 ∠C

2. 2

[similarly bisect

∴∠D= ∠B+∠C

2 ....(1)

, ∠E= ∠C+∠A and ∠F=∠A+∠B

2. 2. ...(2)

, ∠A+∠B+∠C=180 o

....angle sum property of triangle

∴∠B+∠C=180∘ −∠A ...(3)

Similarly, ∠C+∠A=180 o −∠B and ∠A+∠B=180o−∠C ...(4)

∴∠D= 180 o −∠A

2

∠D=90 o− 1 ∠A

2

∠E=90 o− 1. ∠B and ∠F=90 o − 1. ∠C

2 2