43.

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)  

Given: 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.

To Prove: (i) D is the mid-point of AC.

(ii) MD ⊥ AC

(iii)  

Proof: (i) In ∆ACB,
∵ M is the mid-point of AB and MD || BC

∴ D is the mid-point of AC.
| By converse of mid-point theorem
(ii) ∵ MD || BC and AC intersects them
∴ ∠ADM = ∠ACB
| Corresponding angles
But ∠ACB = 90°    | Given
∴ ∠ADM = 90° ⇒ MD ⊥ AC
(iii) Now, ∠ADM + ∠CDM = 180°
| Linear Pair Axiom
∴ ∠ADM = ∠CDM = 90°
In ∆ADM and ∆CDM,
AD = CD
| ∵ D is the mid-point of AC
∠ADM = ∠CDM    | Each = 90°
DM = DM    | Common
∴ ∆ADM ≅ ∆CDM
| SAS Congruence Rule
∴ MA = MC    | C.P.C.T.
But M is the mid-point of AB