<BAC = <BCE, 2|AD| = |ED|

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

<BAC = <BCE, 2|AD| = |ED|

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: Dutch NMO 2022 p4

The post below has been deleted. Click to close.

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

30630 posts

#1 h Nov 17, 2022, 9:19 PM • 1 Y

Y by Mango247

In triangle $ABC$ , the point $D$ lies on segment $AB$ such that $CD$ is the angle bisector of angle $\angle C$ . The perpendicular bisector of segment $CD$ intersects the line $AB$ in $E$ . Suppose that $|BE| = 4$ and $|AB| = 5$ .
(a) Prove that $\angle BAC = \angle BCE$ .
(b) Prove that $2|AD| = |ED|$ .

$[asy] unitsize(1 cm); pair A, B, C, D, E; A = (0,0); B = (2,0); C = (1.8,1.8); D = extension(C, incenter(A,B,C), A, B); E = extension((C + D)/2, (C + D)/2 + rotate(90)*(C - D), A, B); draw((E + (0.5,0))--A--C--B); draw(C--D); draw(interp((C + D)/2,E,-0.3)--interp((C + D)/2,E,1.2)); dot("$A$", A, SW); dot("$B$", B, S); dot("$C$", C, N); dot("$D$", D, S); dot("$E$", E, S); [/asy]$

This post has been edited 1 time. Last edited by nsato, Feb 13, 2023, 6:43 PM

Z K Y

The post below has been deleted. Click to close.

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

carlitosimo2025

1 post

carlitosimo2025

#2 h Jan 11, 2024, 9:17 PM

Y by

$($ $a$ $)$ $Let$ $\angle$ $ACD$ $=$ $\angle$ $BCD$ $=$ $\alpha$ $and$ $\angle$ $CAB$ $=$ $\beta$ $,$ $then$ $\angle$ $CDE$ $=$ $\alpha$ $+$ $\beta$ $.$ $Since$ $E$ $is$ $on$ $the$ $bisector$ $of$ $CD$ $,$ $\angle$ $ECD$ $=$ $EDC$ $=$ $\alpha$ $+$ $\beta$ $,$ $so$ $\angle$ $BCE$ $=$ $\angle$ $ECD$ $-$ $\angle$ $BCD$ $=$ $($ $\alpha$ $+$ $\beta$ $)$ $-$ $\alpha$ $=$ $\beta$ $=$ $\angle$ $BAC$ $,$ $as$ $desired$ $.$

$($ $b$ $)$ $As$ $\angle$ $BAC$ $=$ $\angle$ $BEC$ $,$ $then$ $CE$ $is$ $tangent$ $to$ $($ $ABC$ $)$ $,$ $and$ $by$ $PoP$ $,$ $CE^{2}$ $=$ $EB$ · $EA$ $=$ $4$ · $9$ $=$ $36$
$\Rightarrow$ $CE$ $=$ $6$ $,$ $since$ $\triangle$ $EBC$ $\sim$ $\triangle$ $ECA$ $,$ $by$ $AAA$ $,$ $BC$ $/$ $AC$ $=$ $CE$ $/$ $AE$ $=$ $6$ $/$ $9$ $=$ $2$ $/$ $3$ $.$ $So$ $by$ $angle$ $bisector$ $theorem$ $AC$ $/$ $BC$ $=$ $AD$ $/$ $BD$ $=$ $3$ / $2$ $,$ $and$ $because$ $AD$ $+$ $BD$ $=$ $5$ $\Rightarrow$ $AD$ $=$ $3$ $,$ $BD$ $=$ $2$ $and$ $2$ $AD$ $=$ $6$ $=$ $EC$ $=$ $ED$ $,$ $as$ $desired$ $.$

This post has been edited 1 time. Last edited by carlitosimo2025, Jan 12, 2024, 10:08 PM
Reason: the solution wasnt complete

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a