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

Revision as of 13:07, 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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Difference between revisions of "User talk:Bobthesmartypants/Sandbox"

Revision as of 13:07, 29 September 2013

Bobthesmartypants's Sandbox

@@ Line 1: / Line 1: @@
 ==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 <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>.
+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>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>