Question

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

Answer 1

Given:          ABC is a triangle right angled at B and AB=BC
                      DEFB is square
 To prove:      AE=EC
   Proof:     
                   Since DEFB is a square
                         BD=BF⇒eq.1
                         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              

Answer 2

Step-by-step explanation:

let the square be CMPN

all sides are equal so

CM=MP=PN=CN

ΔABC is isoceles so AC=BC

AN+NC= CM+MB

as CN= CM so.

AN=MB

now consider ΔANP and ΔPMB

AN=MB

∠ANP= ∠PMB= 90

PN=PM

so both triangles are congurent

and AP=PB.