The Total solution for NCERT class 6-12
ABC is atriangle right angled at C. A line through the mid-point M of hypotenuse AB andparallel to BC intersects AC at D. Show that(i) D is the mid-point of AC(ii) MD ⊥ AC(iii) CM = MA = ½ AB
(i) In ΔACB,
M is the midpoint of AB and MD || BC
, D is the midpoint of AC (Converse of mid point theorem)
(ii) ∠ACB = ∠ADM (Corresponding angles)
also, ∠ACB =90°
, ∠ADM =90° and MD ⊥ AC
(iii) In ΔAMD and ΔCMD,
AD = CD (D is the midpoint of side AC)
∠ADM = ∠CDM (Each 90°)
DM = DM (common)
, ΔAMD ≅ ΔCMD[SAS congruency]
AM = CM [CPCT]
also, AM = ½ AB (M is midpoint of AB)