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

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(→‎Bobthesmartypants's Sandbox)
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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>.
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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>.

Revision as of 14:09, 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, $\angle PBA=\angle PDA$. $\Box$

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