Question

I will mark as brainilist for the answer. Solve the second part in detail

Answer 1

In triangle AMC and triangle BMD
AM=BM (As M is the midpoint of AB)
CM=DM (given)
angle AMC = angle DMB ( vertically opposite angle)
therefore , ∆AMC congruent to triangle BMD ( By S.A.S. congruency rule)
this implies , angle MDB = angle ACM ( BY
c.p.c.t.)
BUT angle MDB and angle ACM forms a pair of alternate interior angle
therefore , AC parallel to DB
NOW, angle ACB + angle DBC = 180° ( adj. angle)
90° + angle DBC = 180°
angle DBC = 180°-90°=90°
this implies DBC is a right angle
hence proved