Difference between revisions of "User talk:Bobthesmartypants/Sandbox"
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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>. | ||
− | 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 13:14, 29 September 2013
Bobthesmartypants's Sandbox
First, continue to hit at . Also continue to hit at .
We have that . Because , we have .
Similarly, because , we have .
Therefore, .
We also have that because is a parallelogram, and .
Therefore, . This means that , so .
Therefore, .