Question: $ABD$ is a triangle right angled at $A$ and $AC \perp BD$ then show that $AB^2 = BC \times BD$.
My attempt: So in the beginning, I will show that $ \Delta DAB \sim \Delta DCA$ and continue my proof.
I noticed that I could also take $\Delta DAB \sim \Delta ACD$.
Can anyone please clear my dilemma?
PS: I am a student and while proving Pythagoras Theorem (that's in my course) by similarity of triangles I mistakenly a triangle's name the other way around (as illustrated above) and it led me to trouble and I had to open my book and see what triangle name should I take
While proving the triangles similar in right $\Delta$ keep in mind that first write the common angle then $90^\circ$ angle and then the remaining ones. For e.g. to write $\Delta DAB$ similar to $\Delta ACD$ observe that in both $\Delta \angle D$ is common to both $\Delta's$ and $\angle A = \angle C = 90^\circ$ and the remaining ones are $\angle B$ and $\angle A$ (respectivelty in both the triangles). So we can write $$\Delta DAB \sim \Delta DCA$$ For any triangle like if it is given that $${BA \over PQ}={BC\over RQ}={CA\over RP}$$ to write of similar triangles, see numerator and denominator of first two fractions you will observe that they have B and Q common in numerator and denominator respectively. Similarly first and last have A and P common and last two have C and R common. So you can write $\Delta BAC \sim \Delta QPR$