How do we know for sure these triangles are 3-4-5?

[The Blade Dancer]

I found a maybe-error here. (Probably not)

How do we know for sure these triangles are 3-4-5? They sure look like it, but I thought we shouldn't believe in the pictures so much.

I may be repeating myself (maybe I did a post on this before), but at 4:00, why does 4/PD = PD/9?

[nastya]

Module 2 Week 2 Day 5 Your Turn Explanation Part 3 Mini-Question

Hi @The-Darkin-Blade!

Nice question! Here you have to remember the following fact:

If you draw the height of a triangle, this divides the triangle into two right triangles, each of which is similar to the original one.

Proof:
Let $\angle BAC=\angle BAD={\color{green}\alpha}.$ Then $\angle BCD=\angle BCA=180^{\circ}-\angle ABC-\angle BAC={\color{purple}90^{\circ}-\alpha} \Rightarrow \angle DBC=180^{\circ}-\angle BDC-\angle BCD=180^{\circ}-90^{\circ}-({\color{purple}90^{\circ}-\alpha})={\color{green}\alpha}=\angle BAC=\angle BAD.$ The same way we get that $\angle ABD={\color{purple}90^{\circ}-\alpha}=\angle BCD=\angle BCA.$ All this means that $\bf \triangle ABC\sim {\color{blue}\triangle ADB}\sim {\color{darkorange}\triangle BDC}.$ $\boxed{}$

 
$CE$ is a height in right $\triangle BDE,$ so $\color{darkorange}\triangle CDE$ and $\color{purple}\triangle BCE$ are similar to $\triangle BDE.$

The same thing works with the other triangles:
Let $DE\cap AF=H$ and $BE\cap DF=K.$ Then $DH$ is a height in right $\triangle ADF,$ so $\triangle AHD$ and $\triangle DHF$ are similar to $\triangle ADF.$

In addition, $BE \parallel AH \Rightarrow \triangle BDE\sim \triangle ADH,$ as well as $EK \parallel HF \Rightarrow \triangle KDE\sim \triangle FDH.$

So if one of these triangles is 3-4-5 right triangle, then all the others will also be 3-4-5 right triangles.

 
The same fact that I mentioned before is used at 4:00 in order to get that equality:

 

 
$DP$ is a height in right $\triangle ADC,$ so $\triangle APD$ and $\triangle DPC$ are similar to $\triangle ADC,$ which means that:$$\triangle APD\sim \triangle DPC \Rightarrow$$ $$\Rightarrow \frac{\color{darkorange}PC}{PD}=\text{ratio of similarity}=\frac{PD}{\color{blue}PA}$$ $$\Rightarrow \frac{\color{darkorange}4}{PD}=\frac{PD}{\color{blue}9}$$

[divinedolphin]

https://forum.poshenloh.com/topic/138/question-how-do-you-know-that-they-are-all-3-4-5-triangles