Can someone please help me with these three geometry questions??

Dean drew a large right triangle ABC. He found and labeled the midpoint of the hypotenuse X. He drew a line segment YX parallel to line segment BC. His next step was to draw line segment XZ parallel to the line segment AB. He finally drew line segment BX.

* What is the length of YX in relationship to BC

*What is the area of triangle AYX in relationship to the area of triangle ABC? Explain how you know

* Prove that the tree triangles below are similar to each other

First, prepare a right triangle whose hypotenuse is AC with AB being the vertical side and BC the horizontal side. Then do those drawings for XY and XZ.

Question 1: * What is the length of YX in relationship to BC
Answer: Thales theorem says that YX/BC = AX/AC. Since AX is half of the AC, therefore YX = BC/2.

Question 2: *What is the area of triangle AYX in relationship to the area of triangle ABC? Explain how you know.
Answer: Again, using Thales theorem we will figure out that AY = AB/2 as well. This means that the triangle AYX has vertical and horizontal dimensions that are twice smaller than that of ABC. Since the area of a triangle is found by using a product of the bases: Big Area = 1/2 AB*BC, we get a total of a factor of 4 times smaller area for the smaller triangle. Small Area = 1/2 AY*YX = 1/2 (AB/2)*(BC/2) = 1/2 AB*BC/4 = New Area/4.

Question 3: * Prove that the tree triangles below are similar to each other.
Answer: Proof can be done using similarity postulates. Good luck! (This needs some drawing, so I leave it for you).

As far as #2. Simply follow the explanations I made. That is, following the same method of #1, you would first find: AY = AB/2. That is "AY is half of AB". And then the area is found by multiplying bases! Since both base have sides that are 0.5 of the old triangle's sides. You would get a factor of 0.5*0.5 extra area. That is 0.25 of the old triangle or the quarter of the old one. Mathematically, these are as I described above in the answer section. Hope it makes sense.

Great that you found the answer to #3. That is right!

For question #2, let us call the area of the ABC with A_{1}_{.} And that of AYX with
A_{2}. Then we found that A_{2} is 4 times smaller than A_{1}. That is:

A_{2} = A_{1} / 4.

This factor of one-fourth appeared as a result of the multiplication of the sides, that were twice smaller.

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