Question

In the figure, ABCD is a square whose diagonals AC and
BD intersect at O. The ABEC is a right-angled isosceles
triangle, ZBEC being a right angle. Prove that OBEC is a
square.​

Answer 1

Answer:

given : ABCD is a square

diagonal AC and BD intersect at O

also with side an isosceles triangle BEC right angled at E is drawn

to prove : OBEC is a square

proof :

AC = BD (Diagonals of square are equal )....eq.1

a square is a type of rhombus with opp. sides equal

diagonals of rhombus bisect each other at right angle

so AO= OC...eq.2

BO= OD...eq 3

and angle 1 = 90 degree

from above 3 equations OB= OC

angle 3 = angle 2 ( angles opp. to equal sides are equal )

in triangle BOC

angle 1 + angle 2 + angle3= 180degree ( angles sum prop. of a triangle)

90 degree + 2(angle 3)=180degree

( angle2 = angle3)

2(angle3) = 180degree-90degree= 90degree

angle3= 90/2 = 45 degree

angle 2 = angle3 = 45 degree

in triangle BEC

BE = CE (Given)

so angle CBE = angle ECB ( angles opp. to equal sides are equal)

and angle E = 90 degree

by applying angle sum property of triangle BEC

WE find that angle CBE = angle ECB = 45degree

in triangle BOC and triangle BEC

angle 1 = angle BEC ( 90 degree each)

angle 3 = angle ECB ( 45 degree each)

BC= BC ( common)

by AAS congruence axiom

triangle BOC = triangle BEC

OB= OE (cpct)

now we find that all sides are equal in quad. OBEC and angles are of 90 degree each

so OBEC is a square

hope it helps u..