angle chasing, double angle and equal segments

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

angle chasing, double angle and equal segments

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: Euler Olympiad 2012 P7 [junior]

The post below has been deleted. Click to close.

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

30629 posts

#1 h Sep 24, 2018, 3:38 PM • 1 Y

Y by Adventure10

The angles of the triangle $ABC$ satisfy the condition $2 \angle A + \angle B = \angle C$ . Inside this triangle, the point $K$ is chosen on the bisector of the angle $A$ such that $BK = BC$ . Prove that $\angle KBC = 2 \angle KBA$ .

Z K Y

The post below has been deleted. Click to close.

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

4379 posts

#2 h Sep 25, 2018, 6:01 AM • 2 Y

Y by Adventure10, Mango247

1) Construction: take a segment $AC$ , a point $D$ onto its perpendicular bisector; reflection of $AC$ in $CD$ will cut $AD$ at $E$ ; extend $(AE$ to $B|BE=CE$ , and construction is done. See that $\angle BEC=3\hat{A}, \angle CEA=2\hat{B}\implies 3\hat{A}+2\hat{B}=180^\circ$ or $\hat{A}+\frac{2\hat{B}}{3}=60^\circ$ .
Take now $F\in |AC|$ so that $\angle BFC=60^\circ$ . Finally reflection $K$ of $C$ in $BF$ is incenter of $\triangle BAF$ , because $\angle BFK=\angle AFK=60^\circ, \angle FBK=\angle ABK=\frac{\hat{B}}{3}$ , thus $BK=BC, \angle CBK=2\angle ABK$ and $\angle FAK=\angle BAK$ .

Best regards,
sunken rock

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a