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

okay let's construct a rectangle where the short legs will be this length of a B and the long legs will be this length of CD the first thing I need to do of course is take my straightedge give myself a nice long segment to work from and let's go ahead and start with the CD I like long rectangles rather than tall that's just my preference so first of all there see I'm going to come up and measure with my compass the length of CD there's my construction work I'm going to come down here to see make that construction work and I now know that that is point d well now what I need to do is draw the perpendiculars with my compass so I need to start by giving myself points that are equidistant to see okay so there's two points that are equidistant and we'll start on this side open my compass up and i'm going to start what i like to call the cats i come up now i'm going over to that other point which was equidistant finish off the cat side and now i know if i use my pencil and my straight edge through these two points of intersections that i will create a perpendicular through that point c make it nice and long all right i need to do the same thing over here from those two points of intersection i now need to create my cat's I so I open the compass up past the point D and I get that nice smooth art I'm going to come over to my other point equidistant make that nice smooth arc and again using my pencil on my straight edge I get a perpendicular through d now remember make it nice and long ok so now I have a right angle here and a right angle here my rectangle is very close to being complete but here's my long edge CD now I need to create a be so I need to bring my compass back up and I'm going to measure with my compass a be show my construction work now from see I'm going to make that point and I'm going to label it right there this could be like a bee and keeping the same setting on my compass I'm gonna come over here to the D put the mark and I know that right here this is also a B and these two are now congruent to each other and finally I can take my straightedge connect these points and because these are right angles these should be right angles and because this is CD this should also be CD and I can always check that if I'm not sure by coming down and measuring my CD with my compass and then coming up here from a to a-put bad names that you get the idea and sure enough it's the same length so now these two are congruent to each other and I've created a rectangle

welcome to a lesson on how to construct an altitude of a triangle an altitude is a line segment from a vertex that is perpendicular to the opposite side so this red segment is an altitude because it's a line segment from this vertex to the opposite side and is perpendicular to the opposite side each triangle has three altitudes from each of our techs and the point of concurrency is called the orthocenter so here 0.0 is the point of concurrency with a point of intersection of the three altitudes and this is called the orthocenter now we're going to take a look at constructing an altitude for an acute triangle in an up to triangle and for these constructions we're going to need a compass and a straightedge so for this example we have an acute triangle and we'll construct the altitude from this vertex so the first step is to put the point of the compass on this vertex and open it far enough so that when we swing an arc it will intersect the opposite side in two points and if you want you can extend this side of the triangle so in this example if we swing an arc that look similar to this notice how it intersects the opposite side in two points we have a point of intersection here and we have another point of intersection here now we're going to take the compass and put the point on one of the points of intersection and then open it so that it's more than half the distance from this intersection point to this intersection point and once we do that we'll swing an arc above and below this side of the triangle so it might look something like this we want to make sure we swing them so that they would extend past what we think would be the midpoint of the segment between these two points of intersection and now without adjusting the compass we'll put the point on the other point of intersection here and then swing another arc above and below this side of the triangle and those arcs might look something like this now we're going to take our straightedge and line up the intersection of the arcs here and here with this vertex and this should be our altitude so this red segment would be our altitude because it connects this vertex to the opposite side and is perpendicular to that opposite side now let's take a look at the construction when we have an obtuse triangle if we want to construct the altitude from this vertex here to the opposite side notice how we're going to have to extend the opposite side in order to make a segment that's perpendicular to that side so with our straightedge we'll go ahead and extend this side and then we'll perform the same procedure we'll take the point of our compass put it here on the vertex and then we'll open up the compass large enough so that when we swing an arc it will intersect the opposite side in two points and when we swing an arc it might look something like this and notice how it intersects the opposite side in two points we have one point of intersection here and one point of intersection here next we'll take the compass and put the point on one of the points of intersection and then open up the compass so that extends past half the distance from these two points of intersection and then swing an arc above and below this side of the triangle so if we put the point on this point of intersection and swing an arc above and below it might look something like this and then without adjusting the compass we'll put the point on the other point of intersection and do the same swing an arc above and below the side of the triangle so it might look something like this now if we line up the points of intersection of these arcs with this vertex it should form our altitude so with a straight edge we'll line these three points up they should be linear and construct our altitude and notice how this segment is perpendicular to the opposite side here therefore this should be our altitude I hope you found this helpful you

## Recent Comments