Construct Congruent Angle 2.m4v



let's construct a congruent angle I have angle XYZ it's a fairly large angle the first thing I need to do is give myself a nice straight rate to work from so I'm going to use my straightedge I'm just gonna get myself a nice ray okay so we made ourselves a nice rate of work from this could also be like our endpoint Y here I'm going to take my compass and I need to start from Y open the compass a comfortable amount and draw a nice arc through both rays now I'm going to keep the same compass setting it's really important that I not change that I'm going to come down here put my compass point on my new Y and I need to make a mark that's visibly larger than the other one see how much bigger that is now what I'm going to do is use my compass as a measuring tool and I'm going to measure this arc inside the two rays so I need to really be careful about adjusting my compass and see how I make that nice Construction mark there well I'm going to come down from this point of intersection I'm going to make a similar construction work and now this piece of art and this piece of arc should be the same length what that means is when I use my straightedge and my writing utensil going through this point through the point of intersection I have now created an angle here who is congruent to this angle here and since this is angle XYZ I could also call this angle XYZ because they are the same size they are congruent

Construct Congruent Triangle ASA



our next one is going to be copying the triangle using angle-side-angle congruence use a different one since white is confusing okay so angle-side-angle I still think it's probably easiest to copy the included side first so I'm just going to pick side AC again it doesn't really matter which one you use so we need array longer than AC well the our our triangle rst this time it's going to copy AC down to our a call it point T there so there's our included side which means we need to make an angle at R and an angle at T so a and r have to be congruent and then C and T have to be congruent so we're going to use the construction for copying an angle here kind of like we did in the last one okay so we make an arc from vertex a we're going to place that same arc here at vertex R and we're going to measure the distance between the intersections and place it down below what this angle is going to give us is array that's going to contain the congruent side so I just make it a little bit longer here so I know it's going to cover it so there's our angle and now we need the other angle on this side so we do the same process construct an arc at vertex C make the same arc at vertex T measure the distance between the intersection just a note I made this kind of small so it's a little difficult this compass so your first arc you might want to make a little bit bigger okay but you're measuring this a little smaller then you take the same measurement down here we draw that gray and so here we have angle side angle and our third vertex is right here

Construct Congruent Triangle SAS.m4v



next we're going to copy the triangle using side-angle-side and we can go and just copy this triangle here doesn't really matter so first thing we need to start off with is a side we just start with our array there we'll use some different letters how about L will correspond here with X and then let's go ahead and get at Z again okay so again for the copier segment XZ down onto L and call that n okay so we have one side here now we're going to do is copy an angle at this point we can either copy angle X or angle Z so let's go and copy angle X we're going to use the techniques we learned to copy an angle and we basically put our point of our compass on vertex X and then we make an arc that intersects both rays we copy that arc onto vertex L okay now what we want to do is actually measure the distance between these intersection points okay just want to measure that distance there make sure we're on and then move your compass down to the other intersection and draw an arc that intersects our first arc once you have that you can draw the ray from point L through that intersection so so far we have our side and an angle the last thing to get here is our side so we're going to copy side X Y and put it right down here so there's XY again and we know that that will be the same length let's call that point M and now just connect the segment MN okay so this would be side-angle-side congruence now we have another congruent triangle

(I.46) Construct a square from a given line, Proof



in proposition 46 we proved that with any given line we can always construct a square to begin we are given line a be first with point a and line a B we apply proposition 11 to construct a perpendicular line AC therefore angle BAC is a right angle next with line a B and line AC we apply proposition 3 to make line ad congruent to line a eb now with line a b and point d we apply proposition 31 to construct a parallel line through point D therefore line de is parallel to line a be next with line ad and point B we apply proposition 31 again to construct a parallel line through point B therefore line ve is parallel to line ad since we have two pairs of parallel lines that create shape a de be by definition 6a de be is a parallelogram now since a DB is a parallelogram by proposition 34 the opposite sides are congruent therefore line a B is congruent to line de and line b e is congruent to line ad so now we have line ad is congruent to line a B and line a B is congruent to line de with these two facts we can apply axiom one to show that all three lines are congruent since we also have the line b e is congruent to line ad then by applying axiom one we can show that all four lines are congruent next looking at the parallel lines de and a B we can see that both lines are intersected by line ad then by proposition 29 the sum of the two interior angles on the same side are congruent to two right angles therefore angle B ad plus angle a de are congruent to the sum of two right angles now angle BAC is a right angle which can also be called angle B ad this means that angle a de must also be a right angle since angle B ad plus angle a de are congruent to the sum of two right angles now since a de be is a parallelogram by proposition 34 the opposite angles are congruent therefore angle B ad is congruent to angle de B and angle a de is congruent to angle a be e so by axiom one we can show that all angles are right angles now since a de be is a parallelogram that has all of its sides congruent and all the interior angles are right angles then by definition 6a de b is a square therefore we have proven that with any given line we can always construct a square you

How to construct a kite using a compass and a straightedge



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Dienste. In vandag se video wil ons gaan voort met ons konstrukte vandag ons is
gaan die vlieër met a bou reguit rand en 'n kompas wat ons begin
deur 'n lynstuk te teken en ook te weet dat hierdie vorm of die veelhoeke ons is
Om die vlieër te teken is 'n basiese vlieër dat jy gesien het hoe kinders by parke vlieg
vier jaar sal ek hiermee begin Hierdie lynstuk laat ons die twee etiketteer
eindpunte a en b en laat ons nou vat uit ons kompas gereedskap met die kompas
instrument, laat ons die puntige einde op ons sit begin hier op B en ons moet seker maak
dat die radius van hierdie kompas instrument is stel dat dit meer as die helfte van die lengte is
van 'n B as dit klaar is, laat ons 'n teken trek boog en laat ons dan beweeg om te eindig
Wys a en laat ons dieselfde ding doen By eindpunt a teken ons die tweede boog
en ons soek daardie kruising en ons gaan dit noem
kruis hier C en laat ons nou die kruising wat ons hier by die
onder en laat ons die radius van die kompas en laat ons ons boog teken
hier aan die onderkant en dan laat ons beweeg die puntige punt hierheen om te wees of
eindpunt B en kom ons doen dieselfde ding deur nog 'n boog te voltooi
vir daardie kruising en laat ons se naam hierdie kruispunt hier D alright een keer ons
het ons kruisings wat ons nou nodig het Neem die reguit rand en ons moet
koppel ons eindpunte aan dié aan diegene kruisings goed, laat ons hier begin
deur middel van kruising C met eindpunt met eindpunt B een keer reg
Dit is in plek, laat ons hierdie lyn teken Ons sal dit net ongedaan maak, laat ons dit teken
lyn segment hier Goed en nou gaan ons voort en
beweeg na 'n let se oor na a en kom ons doen en doen dieselfde ding
hier gaan ons van 'n twee gaan om te sien Ons is reg op C en ons is
hier by a en laat ons dit teken kruis hier of teken hierdie lyn
segment sodat hulle hier goed sny laat ons nou koppel
laat ons nou 'n D versamel en van 'n D aansluit Goed daar is ons op en ons
swaai dit terug net 'n bietjie vir D alright En laas maar nie die minste nie, laat ons dit so heroriënteer
dat ons D en B kan koppel, laat ons ongedaan maak wat kom ons hier nader
Goed, sodat daar vir D en D is Kom ons gaan voort en maak dit 'n bietjie skoon
deur ontslae te raak van hierdie boë en dan Ons gaan 'n paar merke byvoeg om te wys
sommige van die eienskappe van a van a vlieër in orde eerste laat ons gaan en
Voeg nog 'n diagonaal by, dit is hier diagonale een wat ek die D een gaan benoem
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genoeg vir 4 D Goed, sodra ons getrek of klaar is
ons konstruksie kom ons gaan voort en Merk hierdie diagonaal hier d2 en dan
Kom ons praat oor sommige van die eienskappe sodra ons ons een keer het
het ons vlieër die twee aangrensende getrek sye gaan kongruent hê
lengtes bo-aan die twee kante of die Daar is ook twee segmente aan die onderkant
gaan kan groei in lengtehoek a hierbo gaan kongruent wees
hoek B en hierdie twee diagonale wanneer hulle sny hulle gaan
sny die skep van 'n loodregte en dit gaan hier vir diagonale wees
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