<C > 90^o if AD/DC=AB/BC, D on AB (1991 Lomonosov Tournament )

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

<C > 90^o if AD/DC=AB/BC, D on AB (1991 Lomonosov Tournament )

G H J

Reveal topic tags

Lomonosov Tournament

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.

30629 posts

#1 h Apr 29, 2021, 3:53 PM • 2 Y

Y by Mango247, Mango247

In the triangle $ABC$ on the side $AB$ , a point $D$ is chosen such that $\frac{AD}{DC}=\frac{AB}{BC}$ . Prove that angle $C$ is obtuse.

This post has been edited 3 times. Last edited by parmenides51, May 20, 2022, 7:48 AM

Z K Y

The post below has been deleted. Click to close.

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

4527 posts

#2 h Apr 29, 2021, 4:20 PM

Y by

Let $E$ be the foot of the angle bisector from $B$ , and let $F$ be the intersection of the perpendicular bisector of $BE$ and line $AC$ . The locus of points $P$ with $\frac{AP}{PC}=\frac{AB}{BC}$ is the circle with center $F$ through $B,E$ . Let this circle intersect line $AB$ again at $D'$ (possibly with $D'=B$ ). For $D'$ to lie on side $AB$ , we must have $\angle FBA<90^{\circ}$ . But $\angle FBA=\angle FBE+\angle EBA=180^{\circ}-\angle C-\frac{1}{2}\angle B+\frac{1}{2}\angle B=180^{\circ}-\angle C$ . Thus, $180^{\circ}-\angle C<90^{\circ}\implies \angle C>90^{\circ}$ , as desired.

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a