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Problem 1: Provide complete derivation of the argument below: (Jr)(Vy) P(r,y) F (Vy)(Jr)P(r,y) Proof:(Jr)(Vy) P(, y) HYP complete the restEnd...

Question

Problem 1: Provide complete derivation of the argument below: (Jr)(Vy) P(r,y) F (Vy)(Jr)P(r,y) Proof:(Jr)(Vy) P(, y) HYP complete the restEnd

Problem 1: Provide complete derivation of the argument below: (Jr)(Vy) P(r,y) F (Vy)(Jr)P(r,y) Proof: (Jr)(Vy) P(, y) HYP complete the rest End



Answers

Write complete proofs for each of the following exercises. Given: $\sphericalangle$ MRK is an exterior angle of $\triangle \mathrm{MAR}$. Prove: $\angle \mathrm{MRK}-\angle \mathrm{M}>0$ (figure not copy)

So based on this diagram, we want to prove that line E f is parallel toe line segment BC, and it is given that a that a e over e b equals F over FC. So, really, what we're proving is the converse of the triangle proportionality serum. Um, which states that if the if a line intersects two sides of a triangle and the and it divides those two sides into proportional segments, then that line must be parallel to the third side of the triangle. So to prove this, um, first of all, it will be helpful. Toe actually, rewrite this proportion in another way, rewrite it as e be over a e equals C over a F. And we're allowed to do that. Um, since they have the same cross products, you're allowed to just take the reciprocal of both sides of a proportion. So now that we did that little manipulation, we can go over to our two column proof. And first of all, let's list are givens. So we'll say, um, he'd be over a e. You know, even though we had to do a little bit of manipulation on the original givens, um, we could just start from here equals FC over f that's given from there. Um, we're actually going to rewrite, um, the numerator er's of this proportion. Um, And what we're gonna do is we'll say e b this segment is the same thing as the larger segment, minus a soon will say a B minus a e over a e equals similarly, um, a c minus, um, a over a f and all we did was substitution there, because we know that is a horrible substitution, but well, because we know from the diagram that, um, line segment E b is just the larger segment minus this segments right here. Um, and we can do that for the other side. From there, we can, um, do a little bit of division so we can separate these fractions as a B over a Yeah, minus. Um, e over a year, which is just gonna become one. Anything over itself is one. And that was just pretty much the reverse of, um, combining like denominators. Right? Um, and we do the same for the other side. So we have a C over a f minus. F over a f is just one and we'll just call that, um algebra. Um because what you're doing is you're just basically distributing the divided by a year. The divided by f t each of the terms. From there, we can add one to both sides to get to cancel out that minus one. So we get a B over a e equals a c over a, uh and that's just the addition. Property of equality. You add one to both sides and the equation is still balanced. Now, from here, you notice that a B, the larger side and a a e the smaller side are proportional to a see again the small, the larger side, and a f of smaller side. And from this, we're gonna try and prove that the smaller triangle E a F. This triangle here is similar to the larger triangle ABC This triangle here and there's one last step. We have to say that angle A is congruent to angle a and that is just the reflexive property. Okay. And from there we can say that Triangle ABC is similar to triangle um A And we knew that by the S. A s similarity serum, which states that if two sides. Um, if the two sides of two different triangles are in proportion to one another, which they are as improved in step four and the included angle is congruent which we showed, and step five, then the triangles are similar. Toe one mother. Now, from there we can show that angle A f is congruent to angle ABC because similar triangles have congruent angles with little shorthand. And lastly, we can finally prove that line. E ah is parallel tow line segment B C. Because if you see here, um, angle a e f an angle ABC are corresponding angles, and if corresponding angles are congruent, then the lines are parallel. So if core angles congruent, then parallel and there we have it. We've proved the converse of the triangle proportionality here, um


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