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

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    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?

  • MOD

    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.

    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:$$\triangle APD\sim \triangle DPC \Rightarrow$$ $$\Rightarrow \frac{\color{darkorange}PC}{PD}=\text{ratio of similarity}=\frac{PD}{\color{blue}PA}$$ $$\Rightarrow \frac{\color{darkorange}4}{PD}=\frac{PD}{\color{blue}9}$$

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