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) D is the mid-point of AC 

(ii) MD ⊥ AC 

(iii) CM = MA = \(\frac{1}{2}\) AB

Answer 1

(i) ∆ABC,

It is given that M is the mid-point of AB and MD || BC. 

Therefore, D is the mid-point of AC. (Converse of mid-point theorem) 

(ii) As DM || CB and AC is a transversal line for them, 

therefore, 

∠MDC + ∠DCB = 180º (Co-interior angles) 

∠MDC + 90º = 180º 

∠MDC = 90º 

∠MD ∠AC 

(iii) Join MC.

In ∆AMD and ∆CMD,

AD = CD (D is the mid-point of side AC) 

ADM = CDM (Each 90º) DM = DM (Common) 

∆AMD ∆CMD (By SAS congruence rule) 

Therefore, AM = CM (By CPCT)

However, AM = \(\frac{1}{2}\) AB (M is the mid-point of AB)

Therefore, it can be said that

CM = AM = \(\frac{1}{2}\) AB