plz answer this question
Answers
Step-by-step explanation:
∵ M is the mid-point of AB.
∴ BM = AM [Given]
(i) In ΔAMC and ΔBMD, we have CM = DM [Given]
AM = BM [Proved]
∠AMC = ∠BMD [Vertically opposite angles]
∴ ΔAMC ≌ ΔBMD (SAS criteria)
(ii) ∵ ΔAMC ≌ ΔBMD
∴ Their corresponding parts are equal.
⇒ ∠MAC = ∠MBD But they form a pair of alternate interior angles.
∴ AC || DB
Now, BC is a transversal which intersecting parallel lines AC and DB,
∴ ∠BCA + ∠DBC = 180°
But ∠BCA = 90 [∵ ΔABC is right angled at C]
∴ 90° + ∠DBC = 180°
or ∠DBC = 180° ∠ 90° = 90°
Thus, ∠DBC = 90°
(iii) Again, ΔAMC ≌ ΔBMD [Proved]
∴ AC = BD [c.p.c.t]
Now, in ΔDBC and ΔACB, we have
∠DBC = ∠ACB [Each = 90°]
BD = CA [Proved]
BC = CB [Common]
∴ Using SAS criteria, we have ΔDBC ≌ ΔACB
(iv) ∵ ΔDBC ≌ ΔACB
∴ Their corresponding parts are equal.
⇒ DC = AB
But DM = CM [Given]
∴ CM = (1/2) DC = (1/2) AB
⇒ CM = (1/2) AB