# Question: Diagonal of Rectangle

## Comment on Diagonal of Rectangle

### Line BD is a straight line .

Line BD is a straight line . If line CE bisects line BD and angle CEB is 90 degreees, then angle CED is equally 90 degrees.

With line BD bisecting angle CDA, angle CDB automatically becomes 45 degrees.

What we have now is a 45:45:90 triangle
1:1:root 2
with the hypotenuse being 12, we have an enlargement factor of root 72 which implies line CE is root root 72.

Simplify root 72 and answer is 8.4 which is answer choice (C).

Your conclusion that angle CDB = 45 degrees is incorrect.

IF angle CDB = 45 degrees, then angle CBD = 45 degrees (since all three angles in triangle BCD must add to 180 degrees).
If angle CDB and angle CBD both euqal 45 degrees, then triangle BCD is an isosceles triangle, which means side CB and side CD must be equal.

HOWEVER, we are told that side CB and side CD are NOT EQUAL. Side CB = 16 and side CD = 12

### so what is the angle of CDB

so what is the angle of CDB AND CBD

### To find those angles, we'd

To find those angles, we'd need to use trigonometry (sine, cosine, etc), which is beyond the scope of the GRE. We'd also need a calculator with trigonometric functions.

That said, I can tell you that angle CDB ≈ 53.13 degrees and angle CDB ≈ 36.87 degrees

### that's great, but why the

that's great, but why the diagonal of a rectangle do not bisect the angle of any edge?? I think we made another problem by this rules that diagonal of a rectangle produce 45 degree as it bisect the angle

### I think I got the point ,

I think I got the point , only diagonal of square bisect the angle

### Close. The diagonal of a

Close. The diagonal of a square AND the diagonal of a rhombus will bisect the angle.

### Is there any trick to solve

Is there any trick to solve this in a shorter time?

### Another option is to use

Another option is to use SIMILAR TRIANGLES (more on similar triangles here: https://www.greenlighttestprep.com/module/gre-geometry/video/872)

I'll compare ∆ABD with ∆EDC (they are similar)

Here's WHY they're similar.

First notice that ∠CED is 90º, since it's on the line with another 90º angle. So, ∆ABD and ∆EDC both SHARE a 90-degree angle.

Next, notice that, since AB || CD, we know that ∠ABD = ∠CDE So, ∆ABD and ∆EDC both SHARE another angle.

Since ∆ABD and ∆EDC already SHARE two angles, they must also share the 3rd angle. So, we know that ∆ABD and ∆EDC are similar triangles.

Let's focus on ∆ABD for a while.
We know that AB = 12
Then we can apply the Pythagorean Theorem to see that BD = 20

Now let's COMPARE corresponding sides.
We'll let x = the length of side EC

First, sides BD and CD are CORRESPONDING sides in the two similar triangles.
So, we'll compare their two lengths: 20 and 12

Second, sides AD and CE are CORRESPONDING sides in the two similar triangles.
So, we'll compare their two lengths: 16 and x

We can now write: 20/12 = 16/x

When we solve this equation for x, we get x = 9.6 (answer choice E)

### please in the case of similar

please in the case of similar triangles, why is it that we are comparing side AD and CE, but not AB and CE

### When working with similar

When working with similar triangles, it's easiest to examine the angles.

We know that ∠BAD = ∠CED (both equal 90 degrees)
We also know that ∠BDA = ∠ECD

Likewise side CE is BETWEEN ∠CED and ∠ECD
As such sides AD and CE are corresponding (so we can compare them)

Your question: Why not AB and CE?
Side AB is BETWEEN ∠BAD and ∠ABD
Side CE is BETWEEN ∠CED and ∠ECD
We know that ∠CED = ∠BAD
HOWEVER, it is not the case that ∠ABD = ∠ECD
Therefore, sides AB and CE are NOT corresponding sides.

Does that help?

Cheers,
Brent

### Next, notice that, since AB |

Next, notice that, since AB || CD, we know that ∠ABD = ∠CDE So, ∆ABD and ∆EDC both SHARE another angle.

I don't get this. Just because they are parallel how can we say these two angles are equal...? Super confused.

### When 2 parallel lines (like

When 2 parallel lines (like the two red lines shown here: https://imgur.com/AssFIWH) are intersected by a transversal (the blue line), we can some pairs of parallel lines.

For more on this, watch: https://www.greenlighttestprep.com/module/gre-geometry/video/858

Does that help?

Cheers,
Brent

Thank you Brent

### Can I make an assumption that

Can I make an assumption that when the diagonal BD is dropped horizontally, it is 4 units greater than the length of AB. So BE+ED=16+4=20. Now for small triangle CED, CD=12, ED=4, so EC can be found with Pyth.Thm. However I got incorrect answer. Please correct me as why this assumption is wrong?

### "...when the diagonal BD is

"...when the diagonal BD is dropped horizontally, it is 4 units greater than the length of AB"

This is not correct. How did you arrive at this conclusion?

### I have a question regarding

I have a question regarding 261(11) in the 2nd edition. I'm really not getting how the book came up with the answer 2 * root 29. Based on my understanding we could use the pytha theorem and calculate BD by squaring 14 + squaring 4 = bd square to get root 212 as a answer which implies 14.56.

What am I doing wrong

### To find the length of BD, we

To find the length of BD, we must create a right triangle.

So, draw a line from point B down to the line AD so that this new line is perpendicular to line AD.

At the point where this new line meets line AD, add a point P.

We now have a right triangle BPD, where BD is the hypotenuse.

PB has length 4, and PD has length 10.

Applying the Pythagorean Theorem, we get: 4² + 10² = BD²
Simplify: 16 + 100 = BD²
Simplify: 116 = BD²
So, BD = √116 = 2√29

Does that help?

Cheers,
Brent