equality of areas

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

equality of areas

G H J

Reveal topic tags

cyclic quadrilateral

geometry proposed

U sviti matematyky

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.

207 posts

#1 h Jul 10, 2008, 1:41 PM • 2 Y

Y by Adventure10, Mango247

In convex quadrilateral $ABCD$ , let $E=AB\cap DC, F=BC\cap AD$ . The angle bisectors of $\angle AED,\angle BFA$ intersect at $K$ . Given that $\angle EKF=90^{\circ}$ , prove that

$[AKB]+[CKD]=[BKC]+[AKD]$ , where

$[...]$ denotes the area of triangle $...$

Z K Y

The post below has been deleted. Click to close.

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

196 posts

Petry

#2 h Aug 11, 2008, 9:39 PM • 2 Y

Y by Adventure10, Mango247

Hello!

The bisector of the angle $\angle BEC$ intersect the lines $BC$ , $DA$ at the points $N$ , $Q$ respectively.
The bisector of the angle $\angle BFA$ intersect the lines $AB$ , $CD$ at the points $M$ , $P$ respectively.
Point $X$ is the midpoint of $AC$ and point $Y$ is the midpoint of $BD$ .
I prove your problem in few steps:
1.The quadrilateral $ABCD$ is cyclic.
2.The quadrilateral $MNPQ$ is a rhombus and the point $K$ is it's center.
3. $K\in XY$
4. $\left[AKD\right] + \left[BKC\right] = \left[AKB\right] + \left[CKD\right] = \frac {\left[ABCD\right]}{2}$ .

1. $\angle EKF = 180^{\circ} - \angle KEF - \angle KFE = 180^{\circ} - \angle KEP - \left(\angle PEF + \angle QFE\right) -$
$-\angle KFQ = 180^{\circ} - \frac {180^{\circ} - \angle B - \angle C}{2} - \left(180^{\circ} - \angle D\right) - \frac {180^{\circ} - \angle A - \angle B}{2} =$
$= ... = \frac {\angle B + \angle D}{2}\Rightarrow \angle EKF = \frac {\angle B + \angle D}{2}$ . Since $\angle EKF = 90^{\circ}\Rightarrow ABCD$ is a cyclic quadrilateral.

2. $\frac {AQ}{QD} = \frac {AE}{ED} = \frac {sin\left(180^{\circ} - \angle D\right)}{sin\left(180^{\circ} - \angle A\right)} = \frac {sin\angle B}{sin\angle C}$ . Similarly $\frac {AM}{MB} = \frac {sin\angle B}{sin\angle C}$ , so $\frac {AQ}{QD} = \frac {AM}{MB}\Rightarrow MQ||BD$ . (1)
Similarly $NP||BD$ . (2)
(1),(2) $\Rightarrow MQ||NP$ . (3)
Similarly $MN||PQ$ . (4)
(3), (4), $MP\bot NQ$ , $\left\{K\right\} = MP\cap NQ\Rightarrow MNPQ$ is a rhombus and $K$ is it's center.

3. $\left\{V\right\} = BX\cap MN$ and $\left\{W\right\} = DX\cap PQ$
$MN||AC$ and $X$ is the midpoint of $AC$ $\Rightarrow V$ is the midpoint of $MN$ (5)
$PQ||AC$ and $X$ is the midpoint of $AC$ $\Rightarrow W$ is the midpoint of $PQ$ (6)
(5), (6), $MNPQ$ -rhombus $\Rightarrow K$ is the midpoint of $VW$ . But $VW||MQ||BD$ , so $K\in XY$ .
The point $K$ lies on the Newton-Gauss line.

4.Construct $AA'\bot XY$ and $CC'\bot XY$ , $A',C'\in XY$ . Since $AX = XC$ $\Rightarrow AA' = CC'\Rightarrow \left[AYK\right] = \left[CYK\right]$ (7)
$KY$ is a median of triangle $KBD$ $\Rightarrow \left[DYK\right] = \left[BYK\right]$ (8)
(7), (8) $\Rightarrow \left[AYK\right] + \left[DYK\right] = \left[BYK\right] + \left[CYK\right]\Rightarrow$
$\Rightarrow \left[AYD\right] - \left[AKD\right] = \left[BKC\right] - \left[BYC\right]\Rightarrow$
$\Rightarrow \left[AKD\right] + \left[BKC\right] = \left[AYD\right] + \left[BYC\right] = \frac {\left[ABCD\right]}{2}\Rightarrow$
$\Rightarrow \left[AKD\right] + \left[BKC\right] = \frac {\left[ABCD\right]}{2}\Rightarrow \left[AKB\right] + \left[CKD\right] = \frac {\left[ABCD\right]}{2}$ .
So $\left[AKD\right] + \left[BKC\right] = \left[AKB\right] + \left[CKD\right] = \frac {\left[ABCD\right]}{2}$ .

Attachments:

Z K Y

The post below has been deleted. Click to close.

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

1327 posts

#3 h Aug 13, 2008, 7:02 AM • 2 Y

Y by Adventure10, Mango247

This problem is coming from the past.

The angle bisectors of $\angle E,\ \angle F,$ are perpendicular each other and their intersection point lies on the line connecting the midpoints $M,\ N,$ of the diagonals $AC,\ BD$ of the given cyclic quadrilateral $ABCD.$

Every point $K,$ lies on the line segment $MN,$ has the property as the problem states (Leon Anne).

Kostas Vittas.

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a