# A square is inscribed in an isosceles right angle so that the square and the triangle have one angle common show that the vertices of the square opposite the vertex of common angle bisect hypotenus

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DEFB is square

To prove: AE=EC

Since DEFB is a square

and....

AB=BC(given)⇒eq.2

By subtracting eq.2 with eq.1, we get

AB-BD=BC-BF

AD=FD⇒eq.3

In ΔADE and ΔEFC

∠EAD=∠ECF (∵ AB=AC)

AF=FC (from eq.3)

∠ADE=∠EFC (adjoining angles of the square)

∴ By ASA congruency rule

ΔEDA is congruent to ΔEFC

by cpct

AE=EC

Hence proved