homeschool girl
Junior Member
- Joined
- Feb 6, 2020
- Messages
- 123
The Question
In the figure below, [MATH]D[/MATH] is a point on segment [MATH]\overline{CE}[/MATH] such that [MATH]\overline{AD}\parallel\overline{BE}[/MATH] and [MATH]A[/MATH] is not on [MATH]\overline{BC}[/MATH] Line segments [MATH]\overline{AD}[/MATH] and [MATH]\overline{BC}[/MATH] intersect at [MATH]P[/MATH].
Can [MATH]\angle CAD=\angle CBE?[/MATH] Explain.
My Answer:
When you have a line that intersects two parallel lines, the 8 angles formed can be split into two groups of four congruent angles, we will extend the lines CB and ED to illustrate this better
[MATH]\overline{CE}[/MATH] is also a transversal of [MATH]\overline{AD}[/MATH] and [MATH]\overline{BE}[/MATH], so [MATH]\overline{PDC}=\overline{BEC}[/MATH].
[MATH]\triangle PCD[/MATH] Shares [MATH]\angle BCE[/MATH] with [MATH]\triangle BCE[/MATH], so all angles on [MATH]\triangle PCD[/MATH] are equal to the angles on [MATH]\triangle BCE[/MATH]
Because [MATH]\triangle PCD[/MATH] is a triangle, [MATH]\angle{PCD}+\angle{PDC}+\angle{CPD}=180[/MATH], and because [MATH]A[/MATH] does not lie on [MATH]\overline{BC}[/MATH], [MATH]\angle{ACD}\neq\angle{PCD}[/MATH]. Assuming that [MATH]\angle CAD=\angle CPD=\angle CBE[/MATH] we get [MATH]\angle{ACD}+\angle{PDC}+\angle{CPD}=180[/MATH], but we just saw that [MATH]\angle{ACD}\neq\angle{PCD}[/MATH], which means that [MATH]\angle{CAD}+\angle{PDC}+\angle{CPD}[/MATH] cannot equal [MATH]180[/MATH], but thats impossable, which must mean that our assumption was wrong and that [MATH]\boxed{\angle CAD\neq\angle CBE}[/MATH]
In the figure below, [MATH]D[/MATH] is a point on segment [MATH]\overline{CE}[/MATH] such that [MATH]\overline{AD}\parallel\overline{BE}[/MATH] and [MATH]A[/MATH] is not on [MATH]\overline{BC}[/MATH] Line segments [MATH]\overline{AD}[/MATH] and [MATH]\overline{BC}[/MATH] intersect at [MATH]P[/MATH].
Can [MATH]\angle CAD=\angle CBE?[/MATH] Explain.
My Answer:
When you have a line that intersects two parallel lines, the 8 angles formed can be split into two groups of four congruent angles, we will extend the lines CB and ED to illustrate this better
[MATH]\overline{CE}[/MATH] is also a transversal of [MATH]\overline{AD}[/MATH] and [MATH]\overline{BE}[/MATH], so [MATH]\overline{PDC}=\overline{BEC}[/MATH].
[MATH]\triangle PCD[/MATH] Shares [MATH]\angle BCE[/MATH] with [MATH]\triangle BCE[/MATH], so all angles on [MATH]\triangle PCD[/MATH] are equal to the angles on [MATH]\triangle BCE[/MATH]
Because [MATH]\triangle PCD[/MATH] is a triangle, [MATH]\angle{PCD}+\angle{PDC}+\angle{CPD}=180[/MATH], and because [MATH]A[/MATH] does not lie on [MATH]\overline{BC}[/MATH], [MATH]\angle{ACD}\neq\angle{PCD}[/MATH]. Assuming that [MATH]\angle CAD=\angle CPD=\angle CBE[/MATH] we get [MATH]\angle{ACD}+\angle{PDC}+\angle{CPD}=180[/MATH], but we just saw that [MATH]\angle{ACD}\neq\angle{PCD}[/MATH], which means that [MATH]\angle{CAD}+\angle{PDC}+\angle{CPD}[/MATH] cannot equal [MATH]180[/MATH], but thats impossable, which must mean that our assumption was wrong and that [MATH]\boxed{\angle CAD\neq\angle CBE}[/MATH]
