:)) in isosceles ∆dek with base dk, ef is the angle bisector of ∠e, m∠def = 43°, and dk = 16cm. find: kf, m∠dek, m∠efd.

We have isosceles ∆DEK 

EF is the angle bisector of ∠E,
 m∠DEF = 43°=  m∠FEK ( the bisector divide the angle in two equal angles)

so m∠DEK = m∠DEF +  m∠FEK = 43 + 43 = 86 

∆DEK  is isoceles triangle (given)
the two base angles are equal m∠EDK =  m∠DKE 
and the sum of angles in a triangle = 180

so m∠DEK  + m∠EDK+ m∠DKE = 180 
86  +  m∠EDK+ m∠DKE =  180 

2  x  m∠EDK = 94 

m∠EDK  = 94 / 2 = 47 

in triangle DEF 
m∠EDF + m∠E FD+m∠FED= 180

47 +m∠E FD + 4 3 = 180

m∠E FD = 180-43-47 =  90

 
Find: KF?
in an isoseles triangle the bisector is in the same time median so 
KF= DK / 2 =  16/ 2 = 8 cm 

Answercounterclockwise rotation about the origin by 180 degrees followed by reflection about the y-axisexplanationthe given figure has verticesrecall that for counterclockwise rotation about the origin, when we rotate figure pqrs counterclockwise through an angle of 180° about the origin, the coordinates will becomenext, we reflect in the y-axis by negating the x-coordinates of the resulting figure to obtain,therefore the correct answer is c