Question

show ∆ ABD =~ ∆ CDA by ASA congruence condition​

Answer 1

ABCD must be a rectangle or a square.

From the figure attached,

AC is a diagonal and it intersects both <A and <C equally into two parts(As we know, keep in mind it's 'equally')

i.e,. <BAC = <DAC = 45°

and, <DCA = <BCA = 45° ( as <A and < C are right angles) ....(i)

From ∆ABD and ∆CDA,

<B = <D (Right-angle)

AB = CD or AD = BC ( Measure of the sides of rectangle)

<BAC = <DCA {Angle) from equation (i)

It's obviously ASA (ANGLE-SIDE-ANGLE) congruence condition.

Hence, ∆ ABD =~ ∆ CDA

Hope this will help, mark me as BRAINLIEST if you are satisfied with the solution.