Originally Posted by
Grandad Hello again f1f2f3Number 3 is a bit more tricky, I think.
The easiest way is to use the formula $\displaystyle \tan (A+B) = \frac{\tan A + \tan B}{1-\tan A \tan B}$, noting that $\displaystyle \tan (\angle EBF) = \tfrac13$, and $\displaystyle \tan\angle EBG = \tfrac12$. Then
$\displaystyle \tan(\angle EBF + \angle EBG)=\frac{\tfrac13+\tfrac12}{1-\tfrac13\cdot \tfrac12}=\frac{\tfrac56}{\tfrac56}=1$
$\displaystyle \Rightarrow \angle EBF + \angle EBG =45^o$
But if you must use similar triangles...
Since $\displaystyle \angle CBH = 45^o$, if we can prove that $\displaystyle \angle EBF = \angle GBH$, then we're there. For then we should have $\displaystyle \angle EBF + \angle EBG = \angle CBH$.
So, in the diagram attached, $\displaystyle J$ is the foot of the perpendicular from $\displaystyle H$ to $\displaystyle BG$. We aim to prove that $\displaystyle \triangle$'s $\displaystyle BJH, BEF$ are similar, and hence that $\displaystyle \angle EBF = \angle JBH$.
Let's assume that the length of the side of any of the squares is 1 unit. So $\displaystyle BG = \sqrt5$ (Pythag.)
Now $\displaystyle \triangle$'s $\displaystyle GJH, GAB$ are similar (Can you supply the reason?)
$\displaystyle \Rightarrow \frac{HJ}{HG}=\frac{BA}{BG}$
$\displaystyle \Rightarrow \frac{HJ}{1}=\frac{1}{\sqrt5}$
and $\displaystyle AG = 2AB \Rightarrow JG = 2JH = \frac{2}{\sqrt5}$
$\displaystyle \Rightarrow BJ = BG-JG = \sqrt5-\frac{2}{\sqrt5}=\frac{5-2}{\sqrt5}=\frac{3}{\sqrt5}=3 HJ$
But $\displaystyle BE = 3EF$
Therefore, since $\displaystyle \angle BJH = \angle BEF =90^o, \triangle$'s $\displaystyle BJH, BEF$ are similar, and hence $\displaystyle \angle EBF = \angle JBH$.
Grandad