Question

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) Dis the mid-point ofAC.
(ii) MD ⊥ AC.
(iii) CM = MA = AB.

Answer 1

Answer:

MD ⊥ AC

D is mid point of AC

AM = CM = AB/2

CM + MA = AB

Step-by-step explanation:

MD ║ BC

=> ∠AMD = ∠ABC

& ∠ADM = ∠ACD

∠ACD = 90°

=> ∠ADM = 90°

=> MD ⊥ AD  

=> MD ⊥ AC ( as D is point on AC)

in Δ AMD & Δ ABC

∠A is common & other two angles are also equal

=> Δ AMD ≅ Δ ABC

=> AM/AB = MD/BC = AD/AC

M is mid point of AB

=> AM/AB = 1/2

=> MD/BC = AD/AC = 1/2

=> MD = BC/2  & AD = AC/2

AD = AC/2 => D is mid point of AC

AM² = DM² + AD²

=> AM² = (BC/2)² + (AC/2)²

=> AM² = (1/4)( BC² + AC²)

CM² = (MD² + CD²)

=> CM² = (BC/2)² + (AC/2)²

=> CM² = (1/4)( BC² + AC²)

=> AM² = CM² = (1/4)AB²

=> AM = CM = AB/2

or CM + MA = AB    ( as MA = CM = BM)