If a point is on the bisector of an angle, then it is equidistant from the sides of the angle.
BD bisects ∠ABC
DE and DF are perpendiculars for AB and AC respectively.
∠EBD = ∠FBD
∠DEB = ∠DFB
BD = BD (Common)
Then, triangles BED and BDF are congruent.
ED = DF
BE = BF
Problem 1 :
If BD bisects the angle ∠ABC
m∠ABD = 4x - 10 and m∠CBD = 7x - 31, find ∠CBD.
Solution :
∠ABD = ∠DBC
4x - 10 = 7x - 31
4x - 7x = -31 + 10
-3x = -21
Dividing by -3 on both sides,
x = 7
Applying the value of x in ∠DBC
∠DBC = 7x - 31
= 7(7) - 31
= 49 - 31
∠DBC = 18
Problem 2 :
If BD bisects the angle ∠ABC,
m∠ABD = 3x + 15 and m∠CBD = 4x + 4, find m∠ ABC.
Solution :
∠ABD = ∠DBC
3x + 15 = 4x + 4
3x - 4x = 4 - 15
-x = -11
x = 11
m∠ ABC = ∠DBC + ∠ABD
= 4x + 4 + 3x + 15
= 7x + 19
= 7(11) + 19
= 77 + 19
m∠ABC = 96
Problem 3 :
Find m∠TSU
Solution :
In triangles TSU and USR.
∠UTS = ∠URS
UT = UR
US = US (Common)
∠UST = ∠USR (By CPCTC)
5z + 23 = 6z + 14
5z - 6z = 14 - 23
-z = -9
z = 9
∠TSU = 5z + 23
= 5(9) + 23
∠TSU = 68
Problem 4 :
Given that 𝑚∠𝑊𝑌𝑍 = 63, XW = 5.7, and ZW = 5.7, then 𝑚∠𝑋𝑌𝑍 = _____________.
Solution :
Using angle bisector theorem of triangle.
𝑚∠𝑊𝑌𝑍 = 63
𝑚∠XYZ = 2(63) => 126
Problem 5 :
Find 𝑚∠QRS
Solution :
∠QRT = ∠SRT
∠QRS = 2(26)
∠QRS = 52
Problem 6 :
Find ∠WXZ
Solution :
∠WXZ = ∠ZXY
2x + 5 = 4x - 25
2x - 4x = -25 - 5
-2x = -30
x = 15
Applying the value of x.
∠WXZ = 2x + 5
∠WXZ = 2(15) + 5
∠WXZ = 35
Problem 7 :
(i) Given that FG = HG and m∠FEH = 56, find m∠GEH.
(ii) Given that EG bisects ∠FEH and GF = √2, find GH.
Solution :
i) m∠FEH = 56 = m∠GEH
(ii) Since EG bisects ∠FEH, GF and GH are congruent.
If GF = √2 then GH = √2
Problem 8 :
EG is angle bisector.
(i) Given that ∠FEG ≅ ∠GEH, FG = HG and FG = 10z – 30, HG = 7z + 6 then find GF.
(ii) Given that GF = GH, m∠GEF = 8a , and m∠GEH = 24, find a
Solution :
Accordingly angle bisector theorem,
FG = 10z – 30 ----(1) HG = 7z + 6 ----(2)
(1) = (2)
10z - 30 = 7z + 6
10z - 7z = 6 + 30
3z = 36
z = 36/3
z = 12
Applying the value of FG = 10z - 30
= 10(12) - 30
= 120 - 30
FG = 90
(ii) m∠GEF = 8a ---(1) m∠GEH = 24 ----(2)
(1) = (2)
8a = 24
a = 24/8
a = 3
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