Given the following:
\green{\angle{ABC}} = Tri_Y^\circ
\purple{\angle{BAC}} = Tri_Z^\circ
\overline{BC} \parallel \overline{DE}
(line BC
is parallel to line DE
)C
and F
lie on a straight line.What is \blue{\angle{DAF}}{?}
What is \blue{\angle{CAE}}{?}
NOTE: Angles not drawn to scale.
\Large{{}^\circ}
The measure of the angles in a triangle sum to 180^\circ
.
\red{\angle{BCA}} = 180^\circ - \green{\angle{ABC}} - \purple{\angle{BAC}}
\red{\angle{BCA}} = 180^\circ - \green{Tri_Y^\circ} - \purple{Tri_Z^\circ} =
\red{Tri_X^\circ}
\blue{\angle{DAF}}
is a corresponding angle to \red{\angle{BCA}}
.
\blue{\angle{DAF}}
= \red{\angle{BCA}} = Tri_X^\circ
\blue{\angle{CAE}}
is an alternate interior angle to \red{\angle{BCA}}
.
\blue{\angle{CAE}}
= \red{\angle{BCA}} = Tri_X^\circ
Given the following:
\overline{AB} \parallel \overline{CD}
(line AB
is parallel to line CD
)\green{\angle{EGB}} = X^\circ
\purple{\angle{AGH}} = X^\circ
\green{\angle{BGH}} = 180 - X^\circ
E
and F
lie on a straight line.What is \blue{\angle{GHD}}{?}
What is \blue{\angle{CHF}}{?}
NOTE: Angles not drawn to scale.
\Large{{}^\circ}
\purple{\angle{AGH}}
forms a linear pair with \green{\angle{BGH}}
.
Therefore \purple{\angle{AGH}} = 180^\circ - \green{Y^\circ} = X^\circ
.
\blue{\angle{GHD}}
and \green{\angle{EGB}}
are complementary angles.
Therefore \blue{\angle{GHD}} = \green{\angle{EGB}}
.
\purple{\angle{AGH}}
and \green{\angle{EGB}}
are opposite angles.
Therefore \purple{\angle{AGH}} = \green{\angle{EGB}} = X^\circ
.
\blue{\angle{GHD}}
and \purple{\angle{AGH}}
are alternate interior angles.
Therefore \blue{\angle{GHD}} = \purple{\angle{AGH}}
.
\blue{\angle{GHD}}
and \purple{\angle{AGH}}
are alternate interior angles.
Therefore \blue{\angle{GHD}} = \green{\purple{AGH}}
.
\angle{GHD} = X^\circ
\blue{\angle{CHF}}
and \purple{\angle{AGH}}
are corresponding angles.
Therefore \blue{\angle{CHF}} = \purple{\angle{AGH}}
.
\angle{CHF} = X^\circ
Given the following:
\green{\angle{BDC}} = ANGLE_2^\circ
\red{\angle{DBE}} = ANGLE_1^\circ
What is \blue{\angle{RAND_SWITCH3 === 0 ? "CHE" : ( RAND_SWITCH3 === 1 ? "GHC" : "DHE" )}} {?}
NOTE: Angles not drawn to scale.
\Large{{}^\circ}
The measure of the angles in a triangle sum to 180^\circ
.
\purple{\angle{BHD}} = 180^\circ - \green{\angle{BDC}} - \red{\angle{DBE}}
\purple{\angle{BHD}} = 180^\circ - \green{ANGLE_2^\circ} -
\red{ANGLE_1^\circ} = \purple{ANGLE_3^\circ}
\blue{\angle{CHE}}
and \purple{\angle{BHD}}
are opposite angles are equal.
Therefore \blue{\angle{CHE}} = \purple{\angle{BHD}} = ANGLE_3^\circ
.
\blue{\angle{CHG}}
\blue{\angle{DHE}}
and \purple{\angle{BHD}}
form a linear pair.
Therefore
\blue{\angle{CHG}}
\blue{\angle{DHE}}
= 180 ^\circ - \purple{ANGLE_3^\circ}
.
\blue{\angle{GHC}}
LABEL.remove();
label([4, 2.5], "\\blue{\\angle{GHC}}=" + (180 - ANGLE_3) + "^\\circ", "above");
\blue{\angle{DHE}}
LABEL.remove();
label([2.5, -0.5], "\\blue{\\angle{DHE}}=" + (180 - ANGLE_3) + "^\\circ", "below");
= 180 - ANGLE_3^\circ
Given the following:
\green{\angle{BDC}} = ANGLE_1^\circ
\red{\angle{AIC}} = 180 - ANGLE_2^\circ
\green{\angle{GCH}} = ANGLE_1^\circ
\red{\angle{FGH}} = 180 - ANGLE_2^\circ
What is \blue{\angle{AJF}} {?}
What is \blue{\angle{IHE}} {?}
NOTE: Angles not drawn to scale.
\Large{{}^\circ}
\purple{\angle{DIJ}}
forms a linear pair with \red{\angle{AIC}}
.
Therefore, \purple{\angle{DIJ}} = 180^\circ - \red{180 - ANGLE_2^\circ}
.
The measure of the angles in a triangle sum to 180^\circ
.
Therefore, \pink{\angle{DJI}} = 180^\circ - \green{\angle{BDC}} - \purple{\angle{DIJ}}
.
\pink{\angle{DJI}} = 180^\circ - \green{ANGLE_1^\circ} - \purple{ANGLE_2^\circ}
= \pink{ANGLE_3^\circ}
\blue{\angle{AJF}}
and \pink{\angle{DJI}}
are opposite angles.
Therefore, \angle{AJF} = \angle{DJI} = ANGLE_3^\circ
.
\purple{\angle{HGC}}
forms a linear pair with \red{\angle{FGH}}
.
Therefore, \purple{\angle{HGC}} = 180^\circ - \red{180 - ANGLE_2^\circ}
.
The measure of the angles in a triangle sum to 180^\circ
.
Therefore, \pink{\angle{CHG}} = 180^\circ - \green{\angle{ACD}} - \purple{\angle{HGC}}
.
\pink{\angle{CHG}} = 180^\circ - \green{ANGLE_1^\circ} - \purple{ANGLE_2^\circ}
= \pink{ANGLE_3^\circ}
\blue{\angle{IHE}}
and \pink{\angle{CHG}}
are opposite angles.
Therefore, \angle{IHE} = \pink{\angle{CHG}} = ANGLE_3^\circ
.