The area of a regular pentagon looks about twice as a big as the pentangle star drawn within it. Is it?
Prove that the internal angle bisectors of a triangle will never be
perpendicular to each other.
Triangle ABC has a right angle at C. ACRS and CBPQ are squares. ST
and PU are perpendicular to AB produced. Show that ST + PU = AB
Correct solutions from Mary of Birchwood High School and Andrei
of School 205 Bucharest have contributed to the edited solution
As triangle DEF is equilateral, all its angles are $60$ .
Angle AEF $= 180 - 60 - c = 120 - c$
Angle BFD $= 120 - b$
Angle EDC $= 120 - a$
From triangle FAE, I calculate angle A:
As triangle ABC is an isosceles Angle ABC $=$ Angle ACB $=
Therefore $180 - (120 - b + a ) = 180 - (120 - a + c )$
Therefore $60 + b - a = 60 + a - c$
Therefore $2a = b + c$
Therefore $a = (b + c)/2$
Angle ABC $= 180 - (120 - b + a ) = 180 - (120 - a + a ) =
Angle ACB $= 180 - (120 - a + c ) = 180 - (120 - a + a ) =
Therefore angle BAC is $60$
Therefore triangle ABC is equilateral.