Triangle Geometry: angle bisector and an altitude

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

Triangle Geometry: angle bisector and an altitude

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: British MO (round 2) 2001

The post below has been deleted. Click to close.

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

523 posts

#1 h Nov 7, 2006, 10:47 PM • 2 Y

Y by Adventure10, Mango247

A triangle $ABC$ has $\angle ACB > \angle ABC$ .
The internal bisector of $\angle BAC$ meets $BC$ at $D$ .
The point $E$ on $AB$ is such that $\angle EDB=90^{o}$ .
The point $F$ on $AC$ is such that $\angle BED=\angle DEF$ .

Show that $\angle BAD=\angle FDC$

Z K Y

The post below has been deleted. Click to close.

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

5167 posts

#2 h Nov 8, 2006, 12:42 AM • 2 Y

Y by Adventure10, Mango247

See the attached diagram for additional notation.

Let $B'=\mathcal{S}_{DE}(B)$ . Then obviously $E-F-B'$ , hence $\angle AFE=\angle CFB'=\gamma-\beta$ .

Next, $\angle EDA=\angle BDA-{\pi\over 2}=(\pi-{\alpha\over 2}-\beta)-{\pi\over 2}={\pi-\alpha\over 2}-\beta={\beta+\gamma\over 2}-\beta={\gamma-\beta\over 2}$ .

Construct $A'=\mathcal{S}_{DE}(A)$ . Then $A'-E-F-B'$ and $\angle A'DE=\angle ADE\implies \angle A'DA=\gamma-\beta$ . As $\angle A'DA=\angle A'FA$ , points $A',D,F,A$ are concyclic, lying on circle $K$ . Since $\triangle A'DA$ is isosceles, its circumcenter lies on its symmetrial axis $ED$ , hence $BB'$ is tangent to $K$ , being perpendicular to its radius. By the theorem on inscribed angle and tangent angle, $\angle FDC=\angle FAD=\angle BAD$ .

Attachments:

Z K Y

The post below has been deleted. Click to close.

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

6194 posts

#3 h Nov 8, 2006, 6:22 AM • 2 Y

Y by Adventure10, Mango247

$D$ is the excenter of $\triangle AFE$ , let $I$ be the incenter of $\triangle AFE$ and $\angle FDC=x$ , then $\angle FDE=90-x$ , $\angle IFD=90=\angle DEI$ , so $\angle FIE=90+x$ , then $\angle FIE=90+\frac{\angle FAE}{2}$ , result follows

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a