# Thread: Similar triangles desperate help

1. ## Similar triangles desperate help

I have a few similar triangles problems. I'm really not sure how to do them. Could you please help me, or at least get me started on how to solve it.
Thank you very much!

2. Hello f1f2f3
Originally Posted by 354732
I have a few similar triangles problems. I'm really not sure how to do them. Could you please help me, or at least get me started on how to solve it.
Thank you very much!
Sorry I haven't long to give you a complete answer, but since you said you are desperate, here's some help for the first two.

1) $y = \sqrt{xz}$

$\Rightarrow y^2 = xz$

$\Rightarrow \frac{y}{x}= \frac{z}{y}$

Can you complete?

2) $\triangle$'s ABM, DEA are similar (Can you see why?)

AM = $\sqrt{a^2 + (\tfrac12b)^2}$ and $\frac{DE}{AD}=\frac{AB}{AM}$

Can you complete?

3. Hello again f1f2f3
Originally Posted by f1f2f3
I have a few similar triangles problems. I'm really not sure how to do them. Could you please help me, or at least get me started on how to solve it.
Thank you very much!
Number 3 is a bit more tricky, I think.

The easiest way is to use the formula $\tan (A+B) = \frac{\tan A + \tan B}{1-\tan A \tan B}$, noting that $\tan (\angle EBF) = \tfrac13$, and $\tan\angle EBG = \tfrac12$. Then

$\tan(\angle EBF + \angle EBG)=\frac{\tfrac13+\tfrac12}{1-\tfrac13\cdot \tfrac12}=\frac{\tfrac56}{\tfrac56}=1$

$\Rightarrow \angle EBF + \angle EBG =45^o$

But if you must use similar triangles...

Since $\angle CBH = 45^o$, if we can prove that $\angle EBF = \angle GBH$, then we're there. For then we should have $\angle EBF + \angle EBG = \angle CBH$.

So, in the diagram attached, $J$ is the foot of the perpendicular from $H$ to $BG$. We aim to prove that $\triangle$'s $BJH, BEF$ are similar, and hence that $\angle EBF = \angle JBH$.

Let's assume that the length of the side of any of the squares is 1 unit. So $BG = \sqrt5$ (Pythag.)

Now $\triangle$'s $GJH, GAB$ are similar (Can you supply the reason?)

$\Rightarrow \frac{HJ}{HG}=\frac{BA}{BG}$

$\Rightarrow \frac{HJ}{1}=\frac{1}{\sqrt5}$

and $AG = 2AB \Rightarrow JG = 2JH = \frac{2}{\sqrt5}$

$\Rightarrow BJ = BG-JG = \sqrt5-\frac{2}{\sqrt5}=\frac{5-2}{\sqrt5}=\frac{3}{\sqrt5}=3 HJ$

But $BE = 3EF$

Therefore, since $\angle BJH = \angle BEF =90^o, \triangle$'s $BJH, BEF$ are similar, and hence $\angle EBF = \angle JBH$.

Hello again f1f2f3Number 3 is a bit more tricky, I think.

The easiest way is to use the formula $\tan (A+B) = \frac{\tan A + \tan B}{1-\tan A \tan B}$, noting that $\tan (\angle EBF) = \tfrac13$, and $\tan\angle EBG = \tfrac12$. Then

$\tan(\angle EBF + \angle EBG)=\frac{\tfrac13+\tfrac12}{1-\tfrac13\cdot \tfrac12}=\frac{\tfrac56}{\tfrac56}=1$

$\Rightarrow \angle EBF + \angle EBG =45^o$

But if you must use similar triangles...

Since $\angle CBH = 45^o$, if we can prove that $\angle EBF = \angle GBH$, then we're there. For then we should have $\angle EBF + \angle EBG = \angle CBH$.

So, in the diagram attached, $J$ is the foot of the perpendicular from $H$ to $BG$. We aim to prove that $\triangle$'s $BJH, BEF$ are similar, and hence that $\angle EBF = \angle JBH$.

Let's assume that the length of the side of any of the squares is 1 unit. So $BG = \sqrt5$ (Pythag.)

Now $\triangle$'s $GJH, GAB$ are similar (Can you supply the reason?)

$\Rightarrow \frac{HJ}{HG}=\frac{BA}{BG}$

$\Rightarrow \frac{HJ}{1}=\frac{1}{\sqrt5}$

and $AG = 2AB \Rightarrow JG = 2JH = \frac{2}{\sqrt5}$

$\Rightarrow BJ = BG-JG = \sqrt5-\frac{2}{\sqrt5}=\frac{5-2}{\sqrt5}=\frac{3}{\sqrt5}=3 HJ$

But $BE = 3EF$

Therefore, since $\angle BJH = \angle BEF =90^o, \triangle$'s $BJH, BEF$ are similar, and hence $\angle EBF = \angle JBH$.

Sorry, I'm still not sure how triangles GJH and GAB are similar.
And also for the first question, I'm still not sure how the two triangles are similar. Could you please explain a little more for me?
Thank you so much!

5. Hello f1f2f3
Originally Posted by f1f2f3
Sorry, I'm still not sure how triangles GJH and GAB are similar. Could you please explain? Thank you!
$\angle G$ is common.

$\angle A = \angle HJG = 90^o$

So the triangles are similar (Equal angles).