Thread: Help in proving these two line segments are congruent?

Originally Posted by Crayola

Hello,

I need help proving BF is congruent to ED.

Givens:
Angle FBC is congruent to Angle EDA
The parallel sides of the parallelogram are congruent

using the fact that ABCD is a parallelogram along with the ASA postulate, you should be able to show triangle FBC congruent to triangle EDA ... then segments BF and ED are congruent because they are corresponding sides.

Originally Posted by skeeter

using the fact that ABCD is a parallelogram along with the ASA postulate, you should be able to show triangle FBC congruent to triangle EDA ... then segments BF and ED are congruent because they are corresponding sides.

Well wouldn't that be for if we are trying to solve for triangle ACD and ABC?

Originally Posted by Crayola

Well wouldn't that be for if we are trying to solve for triangle ACD and ABC?

same result ... make a choice and go w/ it.

Originally Posted by skeeter

same result ... make a choice and go w/ it.

But how would proving triangle ACD and ABC prove that DH and KF are congruent?

Originally Posted by Crayola

But how would proving triangle ACD and ABC prove that DH and KF are congruent?

what are you talking about? your original post said this ...

I need help proving BF is congruent to ED

where are points H and K on your diagram?

Originally Posted by skeeter

what are you talking about? your original post said this ...



where are points H and K on your diagram?

Sorry I got confused. Yes it should've been BF and ED.

recommend you have a look at this link ...

CPCTC means 'corresponding parts of congruent triangles are congruent' and..

Originally Posted by Crayola

Hello,

I need help proving BF is congruent to ED.

Givens:
Angle FBC is congruent to Angle EDA
The parallel sides of the parallelogram are congruent

http://i55.tinypic.com/fvihj8.jpg

Angle FBC is congruent to angle EDA - given.

BC=AD - The parallel sides of the parallelogram are congruent

Angle BCA is congruent to angle CAD - WHY?


From the above we can conclude that:

triangle FBC congruent to triangle EDA. (ASA)

So, BF=ED (corresponding sides on congruent to triangles).

Originally Posted by Also sprach Zarathustra

Angle FBC is congruent to angle EDA - given.

BC=AD - The parallel sides of the parallelogram are congruent

Angle BCA is congruent to angle CAD - WHY?


From the above we can conclude that:

triangle FBC congruent to triangle EDA. (ASA)

So, BF=ED (corresponding sides on congruent to triangles).

Alright, thank you! And for isn't Angle BCA is congruent to angle CAD because of the definition of an angle bisector?

Originally Posted by Crayola

Alright, thank you! And for isn't Angle BCA is congruent to angle CAD because of the definition of an angle bisector?

no

Originally Posted by skeeter

no

Huh?

Originally Posted by Crayola

Huh?

Read carefully!

Parallelogram - Wikipedia, the free encyclopedia

Originally Posted by Also sprach Zarathustra

Read carefully!

Parallelogram - Wikipedia, the free encyclopedia

Oh because of the alternate interior angles....

Originally Posted by Crayola

Oh because of the alternate interior angles....

Indeed!