Question

Please Solve The Question!

Answer 1

Nice question!

Here we are given that

ABCD is a square

=> all angles = 90°

M is midpoint of BC

=> BM = CM

EF |_ DM

=> angle EMD = angle DMF = 90°

Proof :-

Since angle DCM is 90° (angle of a square) and angle MCF is its linear pair,

=> angle MCF + angle DCM = 180°

=> angle MCF + 90° = 180°

=> Angle MCF = 90°

Now, in ∆EBM and FCM

angle EBM = angle MCF (90° each)

BM = CM (Given)

angle EMB = angle FMC (vertically opposite)

=> ∆EBM is congruent to ∆FCM (ASA)

=> EB = FC (c.p.c.t)
and EM = MF (c.p.c.t)

Now,

in ∆DEM and ∆DFM,

EM = MF (shown above)
angle DME = angle DMF (since, DM |_ EF)
DM = DM (common)

=> ∆DEM is congruent to ∆DFM (SAS)

=> DE = DF (c.p.c.t)

Now,

DF = DC + FC

and DC = BC (sides of a square are equal)
and FC = EB (shown above)

=> DF = BC + EB

=> DF = EB + BC

but DF = DE (shown above)

=> DE = EB + BC

Hence Proved ;)