MathDB

similarity of triangles

Source: Romanian District Olympiad, Grade VII, Problem 4

October 4, 2018

geometryPure geometry

Problem Statement

Consider the triangle

ABC

with

\angle BAC>60^{\circ }

and

\angle BCA>30^{\circ } .

On the other semiplane than that determined by

BC

and

A

we have the points

D

and

E

so that

\angle ABE =\angle CBD =\angle BAE +30^{\circ } =\angle BCD +30^{\circ } =90^{\circ } .

Note by

F,H

the midpoints of

AE,

respectively,

CD,

and with

G

the intersection of

AC

and

DE.

Show:
a)

EBD\sim ABC

b)

FGH\equiv ABC

Back to Problems View on AoPS