Answer :
the measure of the smallest angle of ΔDEF is approximately 41.41°.
How to solve the question?
In ΔDEF, DM is a median and M is on EF. Additionally, DM = EF, and DL is an angle bisector of ∠EDF, L is on EF, and m∠DLF = 64°. We need to find the measure of the smallest angle of ΔDEF.
Since DM is a median, it divides EF into two equal parts, EM and MF. Thus, EM = MF = DM/2 = EF/2.
Let x be the measure of ∠EDF. Then, we know that ∠EDM = ∠FDM = 90° because DM is a median.
Using the angle bisector theorem, we know that DL/EL = DF/EF. Since DL is an angle bisector, we also know that ∠DLE = ∠ELF = x/2. Therefore, we have:
DL/EL = DF/EF
DL/(EF/2) = DF/EF
DL = DF/2
Now, we can use the Law of Cosines in ΔDEF to find DF in terms of x:
DF² = DE² + EF² - 2(DE)(EF)cos(x)
DF² = DM² + MF² - 2(DM)(MF)cos(x)
DF² = (EF)²/4 + (EF)²/4 - (EF)²cos(x)
DF² = (EF)²/2 - (EF)²cos(x)
Since DL = DF/2, we have:
DL² = (EF)²/8 - (EF)²cos(x)/4
Using the angle bisector theorem again, we know that EL/FL = DE/DF. Since DL = DF/2, we also know that FL = EF - DL = EF/2. Therefore, we have:
EL/EF - EL/2 = DE/DF
EL/EF - EL/(2DL) = DE/DF
EL/EF - EL/(EF/4) = DE/DF
EL = EF(DE/DF)/3
Now, we can use the Law of Cosines again in ΔDEL to find DE in terms of x:
DE² = DL²+ EL² - 2(DL)(EL)cos(x/2)
DE² = (EF)²/8 - (EF)^2cos(x)/4 + [EF(DE/DF)/3]² - 2(DL)(EF(DE/DF)/3)cos(x/2)
DE² = (EF)²/8 - (EF)^2cos(x)/4 + (EF)²(DE/DF)^2/9 - (EF)(DE/DF)(EF/6)cos(x/2)
Since DM = EF, we have DE = DM - EM = EF/2 - EF/4 = EF/4. Thus, we can substitute this into the equation above and simplify:
(EF/4)²= (EF)²/8 - (EF)^2cos(x)/4 + (EF)^2(DE/DF)²/9 - (EF)(DE/DF)(EF/6)cos(x/2)
(EF)²/16 = (EF)²/8 - (EF)²cos(x)/4 + (EF)²(DE/DF)²/9 - (EF)(DE/DF)(EF/6)cos(x/2)
0 = (EF)²/72 - (EF)²cos(x)/4 + (EF)²(DE/DF)²/9 - (EF)(DE/DF)(EF/6)cos(x/2)
Now, we can substitute DL = DF/2 = (EF/4)/2 = EF/8 and EL = EF(DE/DF)/3 = EF(DE)/(3EF/4) = 4DE/3 into the equation above and simplify:
0 = (EF)²/72 - (EF)²cos(x)/4 + (EF)²(DE/DF)²/9 - (EF)(DE/DF)(EF/6)cos(x/2)
0 = (EF)²/72 - (EF)²cos(x)/4 + (EF)^2(DE/DF)²/9 - (EF/8)(4DE/3)(EF/6)cos(x/2)
0 = (EF)²/72 - (EF)²cos(x)/4 + (EF)²(DE/DF)²/9 - (EF²/72)cos(x/2)
0 = (EF)²/72 - (EF)²cos(x)/4 + (EF)²(DE/DF)²/9 - (EF)²cos(x/2)/18
Simplifying this equation, we get:
cos(x)/4 - cos(x/2)/18 = (EF)²/72 - (EF)²(DE/DF)²/9
Now, we can substitute DE = EF/4 and DF = EF/2 into the equation above and simplify:
cos(x)/4 - cos(x/2)/18 = (EF)²/72 - (EF)²/144
cos(x)/4 - cos(x/2)/18 = (EF)²/144
We know that cos(x) is negative because x is the measure of the smallest angle of ΔDEF, so we can take the absolute value of both sides of the equation:
|cos(x)/4 - cos(x/2)/18| = (EF)²/144
Since 0° < x < 180°, we know that cos(x/2) > cos(x), so we can simplify further:
cos(x/2)/18 - cos(x)/4 = (EF)²/144
Now, we can substitute the given value of ∠DLF = 64° into the equation above and solve for EF:
cos(32°)/18 - cos(128°)/4 = (EF)^2/144
0.0289 - (-0.2113) = (EF)²/144
0.2402 = (EF)²/144
EF = √(0.2402*144)
EF ≈ 4.8044
Finally, we can use the Law of Cosines in ΔDEF to find x:
cos(x) = (DE² + EF² - DF²)/(2(DE)(EF))
cos(x) = (EF²/16 + EF² - EF²/4)/(2(EF/4)(EF))
cos(x) = 3/4
x = arccos(3/4)
x ≈ 41.41°
Therefore, the measure of the smallest angle of ΔDEF is approximately 41.41°.
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