I'm working through Ted Shifrin Linear algebra and when introducing the dot product he motivates it by saying that if P and Q are two points in the plane and the angle $\angle POQ$ is a right angle then the triangles $\triangle OAP$ and $\triangle OBQ$ are similar where A is the projection of $\vec{OP}$ along the x-axis and B is the projection of $\vec{OQ}$ along the y-axis (see image).
Now, I'm not that good in geometry so i got little bit lost in this part. How would one prove that the triangles are similar, which theorems/postulates are used?. Thanks in advance.
As given in question, ∠ POQ = 90.
Thus ∠POB and ∠BOQ are complementary to each other, i.e. ∠POB + ∠BOQ =90
Also, OA ⊥ OB,
Thus, ∠BOP + ∠ AOP = 90
Now, comparing both,
We can prove, ∠BOQ = ∠AOP
Also, ∠OBQ = ∠OAP = 90
As two angles are equal, we can apply AA similarity in the two triangles. Thus, they are similar.