Trapezium problem very nice

Art of Problem Solving

AoPS Online

High School Math Grades 9-12, Ages 13-18

Grades 9-12, Ages 13-18

3 M G

BBookmark VNew Topic kLocked

High School Math Grades 9-12, Ages 13-18

Grades 9-12, Ages 13-18

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

Trapezium problem very nice

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

3254 posts

manlio

#1 h May 17, 2025, 11:00 AM • 1 Y

Y by PikaPika999

Given trapezium ABCD with basis AB and CD parallel. Choose a point E on side BC and a point F on side AD such that AE Is parallel to FC . Prove that DE Is parallel to FB.

This post has been edited 3 times. Last edited by manlio, May 17, 2025, 12:54 PM
Reason: K

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

9100 posts

#2 h May 17, 2025, 8:18 PM • 1 Y

Y by PikaPika999

Given the trapezoid $ABCD\ :\ A(0,0),B(b,0),C(c,d),D(a,d)$ .

Choose $F(\lambda,\frac{d\lambda}{a})$ , then $E(\frac{ab(c-\lambda)}{\lambda (c-b-a)+ab},\frac{bd(a-\lambda)}{\lambda (c-b-a)+ab})$ .

Slope of the line $BF\ :\ m_{BF}=\frac{d\lambda}{a(\lambda-b)}=m_{DE}$ , the slope of the line $DE$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

10681 posts

#3 h May 17, 2025, 8:52 PM • 1 Y

Y by Bekzhankam

Let's suppose $CD<AB$ , then we have an intersection $S$ as in the picture.
We have ${{SD}\over {SC}}={{SA}\over {SB}}, {{SF}\over {SA}}={{SC}\over {SE}}$ hence $SB\cdot SD=SE\cdot SF\implies {{SD}\over {SF}}={{SE}\over {SB}}\implies DE//BF$ .
A similar argument holds when $AB<DC$ .
If $AB=CD$ then $ABCD$ is a pgm and $AECF$ is a pgm. Hence $AF=CE\implies DF=BE\implies DE//BF$ .

Attachments:

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

4404 posts

#4 h May 18, 2025, 5:18 AM

Y by

alexheinis wrote:

Let's suppose $CD<AB$ , then we have an intersection $S$ as in the picture.
We have ${{SD}\over {SC}}={{SA}\over {SB}}, {{SF}\over {SA}}={{SC}\over {SE}}$ hence $SB\cdot SD=SE\cdot SF\implies {{SD}\over {SF}}={{SE}\over {SB}}\implies DE//BF$ .
A similar argument holds when $AB<DC$ .
If $AB=CD$ then $ABCD$ is a pgm and $AECF$ is a pgm. Hence $AF=CE\implies DF=BE\implies DE//BF$ .

If... then $ABCD, AECF$ are isosceles trapezoids, otherwise OK!

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

10681 posts

#5 h May 18, 2025, 6:22 AM

Y by

@sunkenrock: I did mean parallellogram. The parallel sides of $ABCD$ are given to be equal, not the upstanding sides. See the picture below for clarification.

Attachments:

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a