MathDB

<EFD=<ACD+<ECB wanted, points on semicircle (2020 Kyiv City MO 10.5.1 11.4.1)

Source:

September 20, 2020

geometrysemicircleanglesequal angles

Problem Statement

Let

\Gamma

be a semicircle with diameter

AB

. On this diameter is selected a point

C

, and on the semicircle are selected points

D

and

E

so that

E

lies between

B

and

D

. It turned out that

\angle ACD = \angle ECB

. The intersection point of the tangents to

\Gamma

at points

D

and

E

is denoted by

F

. Prove that

\angle EFD=\angle ACD+ \angle ECB