Question

PLEASE HELP!!! :))

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.

Answer 1

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