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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

Terminology

Stage: 4 Challenge Level: Challenge Level:2 Challenge Level:2

Correct solutions from Mary of Birchwood High School and Andrei of School 205 Bucharest have contributed to the edited solution below.

As triangle DEF is equilateral, all its angles are $60$ . Triangle ADB withinscribed triangle DEF

Angle AEF $= 180 - 60 - c = 120 - c$

Similarly

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 $= x$

Therefore $180 - (120 - b + a ) = 180 - (120 - a + c )$
Therefore $60 + b - a = 60 + a - c$
Therefore $2a = b + c$
Therefore $a = (b + c)/2$

Part Two

If $a=b=c$

Angle ABC $= 180 - (120 - b + a ) = 180 - (120 - a + a ) = 60$
Angle ACB $= 180 - (120 - a + c ) = 180 - (120 - a + a ) = 60$
Therefore angle BAC is $60$

Therefore triangle ABC is equilateral.