User talk:Bobthesmartypants/Sandbox

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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User talk:Bobthesmartypants/Sandbox

Bobthesmartypants's Sandbox