Question

Triangle ABC is right angled at C, D is a point on AB and F is a point on AC produced such that CDEF is a parallelogram if angle CEF = 35°, find angle BDE.

Answer 1

hi friend
here's your answer looking for

as we know that,
$de || af$
so,

angle DEC + angle ACE = 180

DEC = 180 - 90 = 90

so,

DEF = DEC + CEF = 90 + 35 = 125

now,

consider the triangle BDE and triangle CEF.

now,

DE = CF ( opp. sides of parallelogram ) ______(1)

angle ACE = angle DEB ( corresponding angle )

so,

angle DEB = 90

so, angle DEB = angle ECF = 90________(2)

now,

as in the question one thing is missing that is D is mid - point or the DE = 1/2 AC

so,

by Converse of midpoint theorem,

BE = CE ________(3)



so, triangle BDE is congruent to triangle CEF.

then,

Angle BDE = angle EFC ( cpct ) _______(4)


now,

angle def + angle EFC = 180

EFC = 180 - 125 = 55

and angle BDE = 55. ( by 4)


hope you are satisfied with my answer