In a square ABCD, F and E are any points on CD and CB respectively such that, AF = DE. Prove that angle EPF = 90 degrees.
Please please help me in this sum. My head is breaking.
given: ABCD is a square. and AF=DE
TPT: ∠EPF=90 deg.
proof: in ΔADF and ΔDCE
AD=CD and hypotenuse AF=DE
therefore by the right angle hypotenuse side congruency
ΔADF and ΔDCE are congruent triangles.
let ∠EDC=θ, by the congruent property ∠DAF will also be θ.
∠ADP=∠ADF-∠PDF=90-θ (since∠PDF is same as ∠EDC)
now in the triangle ADP:
therefore ∠EPF=∠APD=90 degree(opposite angles are equal)