Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "User talk:Bobthesmartypants/Sandbox"

(→‎Bobthesmartypants's Sandbox)
(→‎Bobthesmartypants's Sandbox)
Line 20: Line 20:
  
 
First, continue <math>\overline{AP}</math> to hit <math>\overline{CD}</math> at <math>E</math>. Also continue <math>\overline{CP}</math> to hit <math>\overline{AD}</math> at <math>F</math>.  
 
First, continue <math>\overline{AP}</math> to hit <math>\overline{CD}</math> at <math>E</math>. Also continue <math>\overline{CP}</math> to hit <math>\overline{AD}</math> at <math>F</math>.  
 +
 
We have that <math>\angle PAB=\angle PCB</math>. Because <math>\overline{AB}\parallel\overline{CD}</math>, we have <math>\angle PAB=\angle PED</math>.
 
We have that <math>\angle PAB=\angle PCB</math>. Because <math>\overline{AB}\parallel\overline{CD}</math>, we have <math>\angle PAB=\angle PED</math>.
 +
 +
 
Similarly, because <math>\overline{AD}\parallel\overline{BC}</math>, we have <math>\angle PCB=\angle PFD</math>.
 
Similarly, because <math>\overline{AD}\parallel\overline{BC}</math>, we have <math>\angle PCB=\angle PFD</math>.
 +
 
Therefore, <math>\angle PAB=\angle PED=\angle PCB=\angle PFD</math>. We also have that <math>\angle ADC=\angle ABC</math> because <math>ABCD</math> is a parallelogram, and <math>\angle APC=\angle FPE</math>.
 
Therefore, <math>\angle PAB=\angle PED=\angle PCB=\angle PFD</math>. We also have that <math>\angle ADC=\angle ABC</math> because <math>ABCD</math> is a parallelogram, and <math>\angle APC=\angle FPE</math>.
 +
 
Therefore, <math>ABCP\sim FDEP</math>. This means that <math>\dfrac{FD}{AB}=\dfrac{FP}{AP}=\dfrac{DP}{BP}</math>, so <math>\Delta ABP\sim\Delta FDP</math>.
 
Therefore, <math>ABCP\sim FDEP</math>. This means that <math>\dfrac{FD}{AB}=\dfrac{FP}{AP}=\dfrac{DP}{BP}</math>, so <math>\Delta ABP\sim\Delta FDP</math>.
 +
 
Therefore, <math>\angle PBA=\angle PDA</math>. <math>\Box</math>
 
Therefore, <math>\angle PBA=\angle PDA</math>. <math>\Box</math>

Revision as of 13:09, 29 September 2013

Bobthesmartypants's Sandbox

[asy] path Q; Q=(0,0)--(1,2)--(5,2)--(4,0)--cycle; draw(Q); draw((0,0)--(1.5,1)); label("D",(0,0),S); draw((1,2)--(1.5,1)); label("A",(1,2),N); draw((5,2)--(1.5,1)); label("B",(5,2),N); draw((4,0)--(1.5,1)); label("C",(4,0),S); draw((2,0)--(1.5,1),linetype("8 8")); label("E",(2,0),S); draw((2/3,4/3)--(1.5,1),linetype("8 8")); label("F",(2/3,4/3),W); label("P",(1.5,1),NNE); [/asy]

First, continue $\overline{AP}$ to hit $\overline{CD}$ at $E$. Also continue $\overline{CP}$ to hit $\overline{AD}$ at $F$.

We have that $\angle PAB=\angle PCB$. Because $\overline{AB}\parallel\overline{CD}$, we have $\angle PAB=\angle PED$.


Similarly, because $\overline{AD}\parallel\overline{BC}$, we have $\angle PCB=\angle PFD$.

Therefore, $\angle PAB=\angle PED=\angle PCB=\angle PFD$. We also have that $\angle ADC=\angle ABC$ because $ABCD$ is a parallelogram, and $\angle APC=\angle FPE$.

Therefore, $ABCP\sim FDEP$. This means that $\dfrac{FD}{AB}=\dfrac{FP}{AP}=\dfrac{DP}{BP}$, so $\Delta ABP\sim\Delta FDP$.

Therefore, $\angle PBA=\angle PDA$. $\Box$

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=User_talk:Bobthesmartypants/Sandbox&oldid=57179"