Proving the Cyclic Quadrilaterals in Altitudes Problem | Geometry Help

[lolerlol]

Homework Statement

Let AD be and altitude of triangle ABC where angle A is 90 degrees.

Squares BCX1X2, CAY1Y2 and ABZ1Z2 are drawn outwards from the sides.

Let AX1 meet BY2 in U and AAX2 meet CZ1 in V

Prove that each of the quadrilaterals ABDU, ACDV and BX1UV is cyclic

Homework Equations

The Attempt at a Solution

I'm not sure where to start, but I've been given a clue

If angle BUA = angle BDA, the ABDU is cyclic
Let the point where BY2 meets AC be P
Consider triangles AUP and Y2CP

 

[Ouabache]

Welcome to physics forums! As you may have noticed, this is a great place to discuss ideas and problems you may encounter, in math and science.

On your problem, I would start by re-reading this https://www.physicsforums.com/showthread.php?t=94381. There are many knowledgeable people here, who are willing to steer you towards a successful solution. But first, you need to try to do some work on your problem and show us.

I recommend drawing the problem out as you've described it. A large drawing is useful, since you should see separation of the lines you draw, more clearly. I would also recommend using a pencil with a good eraser and a straight edge for this part.
Next, post what you have drawn as an image.

Can you tell us what a cyclic quadrilateral is?
(hint: if you cannot find it in your text, look it up on the web).

 

[Architeuthis Dux]

Since AD is an altitude, we have angle AUP = angle Y2CP (both are 90 degrees)
Also, angle UAP = angle PCY2 (since they are alternate interior angles)
Therefore, by AA similarity, triangles AUP and Y2CP are similar.
This means that AU/AY2 = AP/AC
But we also know that AP/AC = AB/AY1 (since triangles ACP and ABY1 are similar)
Therefore, AU/AY2 = AB/AY1
This implies that AU/AB = AY2/AY1
Since AX1 and BY2 are parallel, we can use the Intercept Theorem to say that AU/AB = AX1/AY1
Similarly, we can prove that AV/AC = AX2/AZ1
Therefore, by the converse of the Intercept Theorem, we have that BX1UV is cyclic.
Similarly, we can prove that ABDU and ACDV are also cyclic using similar arguments.
Hence, the quadrilaterals ABDU, ACDV and BX1UV are all cyclic.

This is a good start to the proof. However, it would be helpful to include a diagram to better understand the problem and your solution. Additionally, you could also mention that since the angles in a cyclic quadrilateral add up to 360 degrees, by proving that each of the angles in ABDU, ACDV, and BX1UV are equal to 90 degrees, we can conclude that these quadrilaterals are indeed cyclic. Furthermore, you could also mention that since AD is the altitude, it is perpendicular to the base of the triangle, which means that angle BDA is also equal to 90 degrees. This supports your earlier statement that angle BUA = angle BDA, which is necessary for ABDU to be cyclic. Similar arguments can be made for ACDV and BX1UV. Overall, your proof is sound and well thought out.