So this problem we are trying to prove If there So what I'm going to do here has tried to help you guys gain intuition about this problem before going on to the proofing going throughout the blanks here. So ask for the intuition. Let's say here I haven't X and y axis took some corn a plane. I just draw two lights. I'm one one and line to okay. And in the problem, we're given that their slopes are identical to one another here, So let's call this line one. Like to sew the slope off line one here should equal to slope off line too. Well, with the slopes earned exactly the same here. Then the rise of a run Here I make these two triangles here should be exactly the same. Came here. Right. So these two should be course flying triangles. 7 90 degree angle here. And if they're still sort of state here than the course by angle here between the ex assistant, all by itself should also be the same. So how it relates the angle with line here trying easy example where we have a vertical line. Yeah, this is a 90 degree angle carry. That would be your slope that relate to your slope here whereas and I have another vertical line with 90 degree angle. So these two angles Air correspondent, one other. You want to consider that? But nevertheless, they should have, um, very special have the same angles here. Well, then, these three triangles are therefore similar because by angle angle similarly, here, a piece two angles here should also be corresponding as well. Is it sad? About 180 degrees. Yeah, but they are similar trying between the two. Well, but the slopes of the same here and the angle from the excessive access are the same here. Theun intuitively they should be Carol because of the angles were different. Let's say I have a vertical line and something was so like this right then they should intersect. Or if I have two lines here, right, that's very close to Slope Hope. Harry. So something that looks like this If I extend it long enough here, then they will eventually intersect. The only time when they're not going to intersect is when their perilous one other and that comes to the fact that they have the same angle from the X axis there. That's intuition. Problem. Let's look at the proof. So the proof gives ups the fact that the slopes of the two lines are equivalent to one another and they are non vertical, so we can use some trying to make a try and go. You have taken the point on the line and dropping it down to the X axis to vertical line sense. A little trick here to prove our show I'll go here is trying to prove that the two lines are in fact parallel. So we're given that disclosure same and by the definition of slope, to get those two ratios in which the slope of line L is given as BC over a c and the slope off line and is given as e f o r D f here right by the rise of a run. So this would be here in order to get the next step happy, right? Since we know that the slope circle we can set the proportions equal to one another here. So in order to do that, we need to use substitution, right, because if the slopes are the same and we know the values off the slopes here, Then we essentially substituted for the slips. Now, next up here is to rewrite the proportion and here And if you were to rewrite the proportion and here because rewrite it such that we have PC over e f is equal to a C or D f. And when you sell the portion this way here, well, it shows that we are comparing the corresponding sites here. He knows C, B, C and E f or the vertical lines and A c and D f r the faces over the horizontal lines within the triangles that are formed and trying on ABC and travel D E f. So we're using mathematical manipulation thing in order to get this proportion here. Right, So we can revive the proportion family. Now we have the right angles. Concurrence there, Um, so we need to compare the two right angles here so you can see here that the right angles here our angle on a C B care. That's one bright angle here and since all right angles are congruent to one another here, right? We can say that angle. A CB is concluded to d f. Yeah, Okay, so we have that. And the two triangles are similar. A wise that the case, he say, Well, we have the sites being proportional to one another R B C over e f and a c D e f. A right decides our correspond to widely. And we have an angle in between the two sides here. That's congruent. Bye. So we can use SCS similarity here To show that Triangle B A C A B C is similar to triangle D E f and to similar triangles will have corresponding angles. Angles will be concluded here, right? So angle B A C is groomed to angle e d f well, with their corresponding angles. Here You're right. Everything used to correspond the angles. I'll stay over here, just say that the two lines out and on parallel to one over here. Because if I have transfers here with two parallel lines here, their corresponding angles or equivalent to one another in this picture the transfers is the X axis here that completes