Thread: congruent triangles

1. Re: congruent triangles

Originally Posted by kastamonu
An swer is 80.But this is not a tri. question.If we can prove that CDE is isosceles....
Draw EF parallel to CD, F on BC. Show that ABE is similar to CFE by showing that AB/CF = AE/CE .

This shows that angle ECD = angle CEF = angle AEB = 80 degrees

2. Re: congruent triangles

Many Thanks. I also got an email from the author he says "Draw a symmetric of AE according to A. then look for a congruent triangle."

3. Re: congruent triangles

Hello kastamonu,
using your labels.At C a semi-circle is drawn thru A and B meeting AE extended at D and AC extended at H. Note the following
arc AB =60,arc BD =2*40= 80, arc DH = 2*20 =40. Angle BCD = 80 (unknown angle)

Triangle ADC is isosceles.AC =CD (radii). Angle BEA is 80. DEC is 80 (vertical angles)
Triangle EDC is isosceles (base angles both 80- angles DEC and BCD)
DE=DC =AB as specified in given info

4. Re: congruent triangles

Originally Posted by bjhopper
Hello kastamonu,
At C a semi-circle is drawn thru A and B meeting AE extended at D and AC extended at H.
Could you please clarify this construct.

Is C the center of the circle thru A and B?

Also D was used as a label in the original question as a point on BE, you are using the same label for a point on AE.
This is confusing.

Thanks

5. Re: congruent triangles

Hello,
I've been working on this problem with no success. Idea, could you be more explicit in your proof of similarity of the triangles?

Kastomonu, it's a language translation problem. Could you draw a diagram that illustrates what "Draw a symmetric of AE according to A" means?

6. Re: congruent triangles

Originally Posted by johng
Hello,
I've been working on this problem with no success. Idea, could you be more explicit in your proof of similarity of the triangles?

AE/EC = AB/FC turned out to be a bit complicated using properties of 40, 60, 80 triangles, Law of cosines, etc
Here is a much easier method.

BG is the bisector of angle ABE, G on AC. angle BGE = 80 and triangle BGE is isosceles, so let b = BG = BE and a = AB = BC = CA = ED

GE/EC = b/a since BE is a bisector in triangle BCG

BF/FC = b/a since EF is parallel to CD

so GE/EC = BF/FC therefore EF is parallel to BG and angle ECD = angle FEC = angle BGE = 80

Many Thanks.

8. Re: congruent triangles

Originally Posted by Idea
Could you please clarify this construct.

Is C the center of the circle thru A and B?

Also D was used as a label in the original question as a point on BE, you are using the same label for a point on AE.
This is confusing.

Thanks

Go to post 11 for labels. I used these in my semi-circle at C solution.Yes it is BE extended to D not AE extended to D.The semi-circle greatly simplifies solution

9. Re: congruent triangles

Originally Posted by johng
Hello,
I've been working on this problem with no success. Idea, could you be more explicit in your proof of similarity of the triangles?

Kastomonu, it's a language translation problem. Could you draw a diagram that illustrates what "Draw a symmetric of AE according to A" means?
I will draw as soon as possible.

10. Re: congruent triangles

we construct C' such that DC' = BE.if we can show that triangles DC'C and
BEC are congruent then it is ok.

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