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