<A=120^o, median AD perpendicular to side AB (Greece Junior 2011)

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

<A=120^o, median AD perpendicular to side AB (Greece Junior 2011)

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.

30650 posts

#1 h Jul 14, 2019, 4:05 PM • 1 Y

Y by Adventure10

Let $ABC$ be a triangle with $\angle BAC=120^o$ , which the median $AD$ is perpendicular to side $AB$ and intersects the circumscribed circle of triangle $ABC$ at point $E$ . Lines $BA$ and $EC$ intersect at $Z$ . Prove that
a) $ZD \perp BE$
b) $ZD=BC$

This post has been edited 2 times. Last edited by parmenides51, Jul 14, 2019, 4:18 PM
Reason: added Z specification

Z K Y

The post below has been deleted. Click to close.

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

68 posts

e_pi_i

#2 h Jul 14, 2019, 4:14 PM • 2 Y

Y by Adventure10, Mango247

What is point Z in this?

Z K Y

The post below has been deleted. Click to close.

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

30650 posts

#3 h Jul 14, 2019, 4:18 PM • 2 Y

Y by Adventure10, Mango247

thanks for noticing, just added the missing info

Z K Y

The post below has been deleted. Click to close.

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

1915 posts

#4 h Jul 14, 2019, 4:30 PM • 3 Y

Y by Path_to_Almighty, Adventure10, Mango247

Solution

This post has been edited 8 times. Last edited by amar_04, Jul 14, 2019, 4:40 PM

Z K Y

The post below has been deleted. Click to close.

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

34 posts

#5 h Jul 16, 2019, 5:58 AM • 2 Y

Y by Adventure10, Mango247

amar_04 wrote:

Solution

first u are saying $T\in CD$ and then u are saying
$\angle TCD=60^\circ$ . how can this be possible?

This post has been edited 2 times. Last edited by Chotu2004, Jul 16, 2019, 6:00 AM
Reason: typo

Z K Y

The post below has been deleted. Click to close.

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

1915 posts

#6 h Jul 16, 2019, 7:20 AM • 1 Y

Y by Adventure10

Chotu2004 wrote:

amar_04 wrote:

Solution

first u are saying $T\in CD$ and then u are saying
$\angle TCD=60^\circ$ . how can this be possible?

Why??? Why not possible. I think it's clear to you why $ZD\perp BE$ . Now by properties of cyclic quadrilateral you get $\angle BCE=90^\circ$ also you get $\angle CEB=180^\circ-\angle CAB=180^\circ-120^\circ=60^\circ$ . Now let $ZD\cap BE=K$ . Now its easy to see that $DKEC$ is a cyclic quadrilateral. Hence $\angle TDC=60^\circ$ and as $TC=TD$ , then $\triangle TCD$ is an equilateral; triangle .Hence $\angle TDC=\angle DTC=\angle TCD=60^\circ$ .

This post has been edited 4 times. Last edited by amar_04, Jul 16, 2019, 7:26 AM

Z K Y

The post below has been deleted. Click to close.

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

30650 posts

#7 h Apr 22, 2024, 12:59 PM

Y by

a) look for orthocenter
b) look for congruent triangles

This post has been edited 2 times. Last edited by parmenides51, May 3, 2024, 8:41 AM

Z K Y

The post below has been deleted. Click to close.

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

30650 posts

#8 h May 3, 2024, 3:12 PM

Y by

a solution on figure without words made by Mike Nannos elsewhere

Attachments:

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a