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 =AB

(i) In ,

Given that: M is the mid-point of AB and

MD || BC

Therefore,

D is the mid-point of AC (By converse of mid-point theorem)

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

Therefore,

∠MDC + ∠DCB = 180⁰ (Co-interior angles)

∠MDC + 90⁰ = 180⁰

∠MDC = 90⁰

MD is perpendicular to AC

(iii) Join MC

In ΔAMD and ΔCMD,

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

∠ADM = ∠CDM (Each 90⁰)

DM = DM (Common)

ΔAMD ΔCMD (By SAS congruence rule)

Therefore,

AM = CM (By CPCT)

However,

AM = AB (M is the mid-point of AB)

Therefore,

CM = AM = AB