# Perpendicular problem

• Dec 23rd 2010, 05:27 AM
sarahh
Perpendicular problem
In the figure below, A, G, B, F, and C are collinear; DB is perpendicular to BE; DG is perpendicular to AC; and EF is perpendicular to AC. If DB is 20 units long, EB is 10 units long, and EF is 8 units long, how many units long is DG?

I know that the answer is 12 but I'm having trouble justifying why the triangle are similiar, I just don't see the proof of it :(
• Dec 23rd 2010, 05:54 AM
Ackbeet
Are you allowed to use trigonometry? Or is this a purely geometric problem?
• Dec 23rd 2010, 06:04 AM
Quote:

Originally Posted by sarahh
In the figure below, A, G, B, F, and C are collinear; DB is perpendicular to BE; DG is perpendicular to AC; and EF is perpendicular to AC. If DB is 20 units long, EB is 10 units long, and EF is 8 units long, how many units long is DG?

I know that the answer is 12 but I'm having trouble justifying why the triangle are similiar, I just don't see the proof of it :(

As DB is perpendicular to BE, then angles DBG and EBF sum to 90 degrees,
since DBG+DBE+EBF=180 degrees.

Then DBG+BDG=90 degrees and also EBF+BEF=90 degrees.
• Dec 23rd 2010, 06:05 AM
sarahh
I think geometric, I don't see how trig could be applied (I know that if we can prove the triangle similiar then sides lengths would be proportional) but I'm stuck on how to prove the triangles similiar.
• Dec 23rd 2010, 06:10 AM
Ackbeet
Here would be a trig approach: find angle EBF using the inverse sin function, find angle DBG using known facts about angle GBF and angle DBE, then use the sin function to find DG.
• Dec 23rd 2010, 06:19 AM
sarahh
Ahh that's even more confusing ackbeet :( Not sure how to do that since we don't know any other angle measures--those angles could be anything!
Archie--well sure it's obvious that the rest of the angles sum to 90 LOL But how do I prove the triangles similiar? AA, SAS.. that type of thing. What's the proof that is..
• Dec 23rd 2010, 06:53 AM
Wilmer
Read Archie's post again...he's spot on!

Let k = angle EBF, then angle BEF = 90-k ; OK?

Since angle DBE = 90 (DB perpendicular to EB), then angle DBG = 180-90-k = 90-k ; OK?

Leaves angle BDG = k ; so triangles similar; OK?

Edit...thanks Soroban; but I beat you by .0001333... second !
• Dec 23rd 2010, 06:53 AM
Soroban
Hello, sarahh!

Quote:

$\displaystyle \text{In the figure below, }A, G, B, F,\text{ and }C\text{ are collinear.}$
$\displaystyle DB \perp BE,\;DG \perp AC,\; EF \perp AC.$

$\displaystyle \text{If }DB\text{ is 20 units long, }EB\text{ is 10 units long, and }EF\text{ is 8 units long,}$
. . $\displaystyle \text{how many units long is }DG?$

$\displaystyle \text{I know that the answer is 12,}$
$\displaystyle \text{but I'm having trouble proving the triangles similiar.}$
Code:

  D *     |@ *     |    *                      * E     |        *                  *|     |          * 20        10 * |     |              *          *  |8     |                *      *  |     |                    *  * @  | * - * - - - - - - - - - - - * - - * - * A  G                      B    F  C

Let $\displaystyle \,\theta \,=\,\angle EBF.$

At point $\displaystyle \,B\!:\;\angle GBD + \angle DBE + \angle EBF \:=\:180^o$

That is: .$\displaystyle \angle GBD + 90^o + \theta \:=\:180^o \quad\Rightarrow\quad \angle GBD \:=\:90^o - \theta$

Hence: .$\displaystyle \angle GDB \,=\,\theta$

Right triangles $\displaystyle DGB$ and $\displaystyle BFE$ have a common acite angle, $\displaystyle \,\theta.$

. . Therefore: .$\displaystyle \Delta DGB \,\sim\,\Delta BFE$

• Dec 23rd 2010, 06:59 AM
sarahh
Wow, that's something I never learned before to prove similiar triangles.. I thought it has to be one of those "congruences" type proofs, like HL... I learned something today :) For hours I was playing around with DG and EF being parallel looking for corresponding angles or something

EDIT: How do you conclude angle GDB = theta Soroban? What about the 90 - theta?

Wilmer--same thing sort of--why is k involved in the equation of DBG saying that it's 90 - k. Why does it have to be k? We don't know anything about that triangle having a "k" angle. Hmmm I thought I got it too :(
• Dec 23rd 2010, 07:30 AM
Wilmer
Quote:

Originally Posted by sarahh
EDIT: How do you conclude angle GDB = theta Soroban? What about the 90 - theta?

Wilmer--same thing sort of--why is k involved in the equation of DBG saying that it's 90 - k. Why does it have to be k?

Not sure HOW to "explain" here by typing...but that's pretty "basic" stuff...

A triangle's 3 angles total 180 degrees, right?
IF a right triangle, then 1 angle = 90 degrees; so other 2 angles total 90 degrees, right?
So if we let 1 of those 2 angles = k (or whatever variable that turns you on!),
then the other angle HAS TO equal 90 - k, right?
• Dec 23rd 2010, 08:33 AM
Quote:

Originally Posted by sarahh
Wow, that's something I never learned before to prove similiar triangles.. I thought it has to be one of those "congruences" type proofs, like HL... I learned something today :)

For hours I was playing around with DG and EF being parallel looking for corresponding angles or something.

EDIT: How do you conclude angle GDB = theta Soroban? What about the 90 - theta?

Wilmer--same thing sort of--why is k involved in the equation of DBG saying that it's 90 - k. Why does it have to be k? We don't know anything about that triangle having a "k" angle. Hmmm I thought I got it too :(

In that case, sarahh,

if you imagine rotating triangle EBF anticlockwise about the point B
until E is on the other triangle's DB side,
then EF will be parallel to GB,
BF will be parallel to GD
and we can see that angle BEF equals angle DBG
and that angle FBE equals angle BDG.

The triangles are right-angled, so both triangles have the exact same angles,
hence they are similar.
• Dec 23rd 2010, 07:36 PM
sarahh
Thanks all, it officially makes sense now :) Happy holidays!