Triangle: Given that angle A congruent to angle BDE and....

[jakeg1116]

I have been stuck on this problem for a while

Given: ?A??BDE; AB x DE = AD x BD

Prove: AB x CD = AD x BC

all I really need is how to prove that ?BAD=?EDC and I can do the rest, here is what I have so far:
I have proven that Triangles ABD and BDE are similar, and I know that ?BCA=?ECD
I have written and tried so many substitutions and just can't seem to get on the right track
it makes me feel pretty stupid, because I have been stuck on this for so long
help would be greatly appreciated

 

[pka]

Re: Triangle help

Your are given \(\displaystyle m\left( {\angle DAB} \right) = m\left( {\angle BDE} \right)\;\& \,m\left( {\angle BDC} \right) = m\left( {\angle BDE} \right) + m\left( {\angle EDC} \right)\).
By the exterior angle theorem you know that \(\displaystyle m\left( {\angle BDC} \right) = m\left( {\angle ABD} \right) + m\left( {\angle BAD} \right)\).

 

[jakeg1116]

Re: Triangle help

so ?ABD=?EDC which i did get using substitution meaning that 2?EDC=?ABC,
i am still at a loss as to how i can get to ?BAD=?EDC. i am probably missing the obvious, but more help would be great

 

[jakeg1116]

bump, please help, I have been stuck for weeks

 

[Deleted member 4993]

I have been stuck on this problem for a while

Given: ?A??BDE; AB x DE = AD x BD

Prove: AB x CD = AD x BC

all I really need is how to prove that ?BAD=?EDC and I can do the rest, here is what I have so far:
I have proven that Triangles ABD and BDE are similar, and I know that ?BCA=?ECD
I have written and tried so many substitutions and just can't seem to get on the right track
it makes me feel pretty stupid, because I have been stuck on this for so long
help would be greatly appreciated

The reason I could not help you, because the special characters that you used above is not displaying properly on my monitor - comes up as little squares. Please describe those some other way.

 

[jakeg1116]

I have been stuck on this problem for a while

Given: angleA is congruent to angleBDE; AB x DE = AD x BD

Prove: AB x CD = AD x BC

all I really need is how to prove that angleBAD=angleEDC and I can do the rest, here is what I have so far:
I have proven that Triangles ABD and BDE are similar, and I know that angleBCA=angleECD
I have written and tried so many substitutions and just can't seem to get on the right track
it makes me feel pretty stupid, because I have been stuck on this for so long
help would be greatly appreciated

Is this better?

 

[Deleted member 4993]

I have been stuck on this problem for a while

Given: angleA is congruent to angleBDE; AB x DE = AD x BD

Prove: AB x CD = AD x BC

all I really need is how to prove that angleBAD=angleEDC

Is the above statement correct?

You can show (pka did that above) that

angle ABD = angle EDC

and I can do the rest, here is what I have so far:
I have proven that Triangles ABD and BDE are similar, and I know that angleBCA=angleECD
I have written and tried so many substitutions and just can't seem to get on the right track
it makes me feel pretty stupid, because I have been stuck on this for so long
help would be greatly appreciated

Is this better?

 

[Mrspi]

I have been stuck on this problem for a while

Given: ?A??BDE; AB x DE = AD x BD

Prove: AB x CD = AD x BC

all I really need is how to prove that ?BAD=?EDC and I can do the rest, here is what I have so far:
I have proven that Triangles ABD and BDE are similar, and I know that ?BCA=?ECD
I have written and tried so many substitutions and just can't seem to get on the right track
it makes me feel pretty stupid, because I have been stuck on this for so long
help would be greatly appreciated

You are GIVEN that <A = <BDE.

You are also given that AB*DE = AD*BD. From this fact, you can write a proportion:

AB/AD = BD/DE (note that if you find the product of the means and extremes of this proportion, you get the above given statement)

Now...triangle ABD ~ triangle DBE (because two pairs of corresponding sides are proportional and the included angles are equal...this is sometimes known as the "S-A-S similarity theorem.")

Because the triangles are similar, ALL of the corresponding angles are congruent. Specifically, <ABD is congruent to <DBE.

That means BD is a BISECTOR of <ABC.

Now, we get to use a theorem that slipped my mind until today...I've thought and thought about your problem! Here's the theorem:

The bisector of an angle in a triangle divides the opposite side into segments that are proportional to the other two sides.

Since BD is a bisector of angle B in triangle ABC, and forms segments AD and DC, AD/DC = AB/BC.

Multiply the means and extremes of that proportion together; the products are equal by the "means-extremes theorem":

AD*BC = DC*AB

There ya go!

You will make yourself crazy by trying to prove that all three triangles are similar to each other, because they JUST PLAIN AREN'T. And I guess the moral of this story is that it is a good idea to review the theorems you've learned recently...these seemingly-impossible proofs are often so difficult because one doesn't take advantage of new tools.

 

[jakeg1116]

thank you so much, I must have completely forgotten that theorem, because I have been retaking this class online from 4 years ago to improve my B (AIMS Tuition Waiver), and couldn't find anything like it in the online lesson. (I KNEW there was some relation with the segments created by the bisector, and couldn't remember it exactly, so i tried to move on) Thanks again.