Question

In ΔDEF, DM is a median, M ∈ EF, and DM = $\frac{1}{2}$ EF. DL is an angle bisector of ∠EDF, L ∈ EF, and m∠DLF = 64°. Find the measure of the smallest angle of ΔDEF.

Answer 1

The two isosceles triangles formed by the median line indicates that

ΔDEF is a right triangle.

Response:

• The measure of the smallest angle of ΔDEF is 19°

Which method can be used to find the smallest angle?

The given parameters are;

DM is a median of ΔDEF

M ∈ EF

$DM = \mathbf{ \frac{1}{2} \cdot EF}$

DL is an angle bisector of ∠EDF

Required:

The measure of the smallest angle of ΔDEF

Solution:

ΔDME and ΔFMD are isosceles triangles by definition of isosceles

triangles.

∠MDF = ∠MFD base angles of an isosceles triangle

∠MDE = ∠MED base angles of an isosceles triangle

∠MDF + ∠MFD + ∠MDE + ∠MED = 2·∠MDF + 2·∠MED = 180° by angle sum property of a triangle

∠MDF + ∠MED = 90°

∠DLF = ∠MED + ∠LED by exterior angle theorem

∠MED + ∠LDE = 64° by substitution property of equality

∠EDF = ∠MDF + ∠MED by angle addition property

• $\angle LDE = \dfrac{\angle EDF}{2} = \dfrac{\angle MDF + \angle MED}{2} = \dfrac{90^{\circ}}{2} = \mathbf{ 45 ^{\circ}}$

∠MED = 64° - ∠LDE

∠MED = 64° - 45° = 19°

∠MFD = 90° - 19° = 71°

∠EDF = 90°

In triangle ΔDEF, ∠MED = 19°, ∠MFD = 71°, ∠EDF = 90°

• The measure of the smallest angle in ΔDEF = 19°

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