Eg 7 of exercise 8.2 of …
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Gaurav Seth 3 years, 10 months ago
ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that
(i) D is the mid-point of AC
(ii) MD ⊥ AC
(iii) CM = MA = ½ AB
Solution:
(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)
Hence, CM = MA = ½ AB
0Thank You