Forever Lost 55 Report post Posted May 5, 2011 I'm having trouble figuring this out right off hand, so I figured I'd see if anyone could explain it, or give some insight that I could use on making a two-column proof for this. Here's the info: Quote Share this post Link to post Share on other sites
Misaki-chan 164 Report post Posted May 5, 2011 O.o What the...? Too... hard... brain.... overload..... *explodes* Sorry I'm no help. ^.^; You should ask Aeyra or Cure-kun. They're smart in math. Quote Share this post Link to post Share on other sites
Aeyra 260 Report post Posted May 5, 2011 Use parallel lines and vertical angles and the answer should come right out. Sorry, in a rush. Longer explanation layer. Quote Share this post Link to post Share on other sites
Detective Rohit 38 Report post Posted May 5, 2011 Quite easy: I dont know how you proove, but here is the proof Before that take the point of intersection of the line CG and EA as.. Say point M In ABDE is a //gram, CD is // to EM and DF is // to CG In CDFG is a //gram, CD is // to FG and DE is // to BA Therefore, EM is // to FG and CM is // to BA (The Trick and it means that FGCM is a //gram too) Now Angle G = Angle M (FG is // to EM) Now Angle M = Angle A (CM is // to BA) Hence Angle G = Angle A. Quote Share this post Link to post Share on other sites
Kaitou Kid Legendary Thief 197 Report post Posted May 5, 2011 Eh... what do you mean by two column proof? Well if you just want explanations, i can give a fairly detailed one: First part Given: ABDE and CDFG are parallelograms --> BD//AE (BD parallel to AE *from diagram*) and CD//FG Since C lies on line BD, BD//CD From the two statements above, we know that BD//AE//CD//GF --(1) Second part Given: ABDE and CDFG are parallelograms --> DF//CG and BA//DE from diagram Since E lies on DF, DE//DF From these two statements above, we know that DE//DF//CG//BA --(2) We need to prove: angle G is the same as angle A, which means this - PROVE angle CGF = angle BAE from equation (2), we know that CG//BA from equation (1), we know that GF//AE Thus, using rule of corresponding angles CGF ~= BAE Quote Share this post Link to post Share on other sites
tengaku squared 291 Report post Posted May 6, 2011 If ABDE and CDFG are parallelograms, you could do it like this: 1) ABDE and CDFG are parallelograms. (given) 2) Angle A is congruent to Angle D. (opposite angels of a parallelogram are congruent) 3) Angel D is congruent to Angle G. (same as 2) 4) Angel A is congruent to Angle G. (If two angles are congruent to the same angle, they are congruent) Quote Share this post Link to post Share on other sites
Aeyra 260 Report post Posted May 6, 2011 *meh* YOU GUYS STOLE MY EXPLANATION!!!! Sorry... I'm having a hard time getting onto DCW other than school hours and five minutes in the morning... Therefore I can't post very long things. Quote Share this post Link to post Share on other sites
Tsukiko 73 Report post Posted May 6, 2011 *frozen* Quote Share this post Link to post Share on other sites
Forever Lost 55 Report post Posted May 6, 2011 Thanks, ya'll, that was really helpful. Quote Share this post Link to post Share on other sites
Balthazar Manfredie 226 Report post Posted May 17, 2011 i never did like geometry, but had to learn it Quote Share this post Link to post Share on other sites