You need to know every postulate and theorem from Socrates postulates in Chapter 1 thru Chapter 5... not just 14 lousy theorems.
MEMORIZE?? You don't have to memorize ANYTHING if you understand the concepts.
For example, the "two non-parallel lines in the same plane" problem from the other day. We used the indirect proof technique to prove that a third line must intersect at least one of them. We stated the negation and assumed that the third line intersected neither. We then stated that the third line must be parallel to both. If that were true, then the two original lines must be parallel to each other, because we knew that "if two lines in the same plane are parallel to a third line, then they are parallel to each other."
We didn't prove the case because we memorized the theorem... we simply UNDERSTAND GEOMETRY.
Gotsk it? Prove it!
As we say in Algebra "the likelihood that you can use your bibles on the unit test DECREASES as the number of requests to use it INCREASES."
I am confused on the proof for number 17 on the Unit 5 chapter test... I proved that triangle QPB is congruent to QPA and that QTA is congruent to QTC...but I am struggling to figure out the next step to proving QC is equal to QB
I don't really understand #8 on the Chapter Test.. Does it mean that PB would bisect <B and <PBC would be half the measure of <ABC? I have a hunch this is correct but I don't know why..
I'm starting to see something but don't know what to do with it.. That <QTC and <QPB are equal in measures and I'm not sure if that means QC=QB because the triangles QAC and QAB are not congruent..
We actually discussed this in class the other day...
My view - We have the perp bisectors as stated, so QB cong QA and QA cong QC so trasitive says QB cong QC.
Mr. Annoying's view - Since you have the perp bisectors, you can create tri-ABC and Q must be the circumcenter. Since QA,QB,&QC are radii, you gotsk yourself congruence.
OHHH! NOW I SEE! I should remember to use all the postulates & theorems & ways of congruence we learned.. So Triangle QAT and QTC are congruent because of SAS and by CPCTC QA is congruent to QC? And then transitive?
For #8, point P is equidistant from all three sides, so it must be the incenter. The incenter is formed by angle bisectors. Draw it and see. <PBC must be half of <B (aka <ABC).
You need to know every postulate and theorem from Socrates postulates in Chapter 1 thru Chapter 5... not just 14 lousy theorems.
ReplyDeleteMEMORIZE?? You don't have to memorize ANYTHING if you understand the concepts.
For example, the "two non-parallel lines in the same plane" problem from the other day. We used the indirect proof technique to prove that a third line must intersect at least one of them. We stated the negation and assumed that the third line intersected neither. We then stated that the third line must be parallel to both. If that were true, then the two original lines must be parallel to each other, because we knew that "if two lines in the same plane are parallel to a third line, then they are parallel to each other."
We didn't prove the case because we memorized the theorem... we simply UNDERSTAND GEOMETRY.
Gotsk it?
Prove it!
As we say in Algebra "the likelihood that you can use your bibles on the unit test DECREASES as the number of requests to use it INCREASES."
Ca-peesh?!
I'm going out for a geogebra holiday party... see y'all later!
ReplyDeleteSo do we get to use the bibles? :(
ReplyDeleteI'll use some inductive reasoning and guess no..
ReplyDeleteUmm question...
ReplyDeleteI am confused on the proof for number 17 on the Unit 5 chapter test...
I proved that triangle QPB is congruent to QPA and that QTA is congruent to QTC...but I am struggling to figure out the next step to proving QC is equal to QB
Will we be able to use our bibles?
ReplyDeleteI don't really understand #8 on the Chapter Test.. Does it mean that PB would bisect <B and <PBC would be half the measure of <ABC? I have a hunch this is correct but I don't know why..
ReplyDeleteNice link, Justin..
ReplyDeleteI have NO idea what to do on #17 or where to start.. I can't see any path that I could take to prove QC=QB..
ReplyDeleteI'm starting to see something but don't know what to do with it.. That <QTC and <QPB are equal in measures and I'm not sure if that means QC=QB because the triangles QAC and QAB are not congruent..
ReplyDeleteMr.C when you get back from your Geogebra party, would you please answer my questions? I'll check for the answers soon or tomorrow morning.
ReplyDeletepg 345 #17...
ReplyDeleteWe actually discussed this in class the other day...
My view - We have the perp bisectors as stated, so QB cong QA and QA cong QC so trasitive says QB cong QC.
Mr. Annoying's view - Since you have the perp bisectors, you can create tri-ABC and Q must be the circumcenter. Since QA,QB,&QC are radii, you gotsk yourself congruence.
ta-da!
WAIT-- How do you know that QA & QB and & QC are all congruent?
ReplyDeleteIt doesn't state anywhere that the two triangles are isosceles...
ReplyDeleteThe perp bisectors create isosceles triangles with an assist from SAS, my dear pamacicos!
ReplyDeleteOhh.. I see how triangle QPB and QPA are congruent... But how do you get to QC being congruent to QA?
ReplyDeleteOHHH! NOW I SEE! I should remember to use all the postulates & theorems & ways of congruence we learned.. So Triangle QAT and QTC are congruent because of SAS and by CPCTC QA is congruent to QC? And then transitive?
ReplyDeleteI misspelled transitive... re-read the two views above...
ReplyDeleteMr. Annoying's view of the circumcenter was quite an interesting "look"
I read it, but does my explanation make sense?
ReplyDeleteAnd I'm still confused on #8..
Mr.C?
ReplyDeleteFor #8, point P is equidistant from all three sides, so it must be the incenter. The incenter is formed by angle bisectors. Draw it and see. <PBC must be half of <B (aka <ABC).
ReplyDeleteOHHHH ok thanks!
ReplyDeletegoodnight... I'll be looking for my voice...
ReplyDeleteGood night & good luck finding it :)
ReplyDelete