Please solve Q96

Please solve Q96 96. ABC is a triangle right angled at C and M is mid-point of hypotenuse AB. Line drawn through M and parallel to BC intersects AC at D. Show that : (i) D is mid-point of AC. (ii) MD is perpendicular to AC.

(i) In ΔABC,

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 ΔAMD and ΔCMD,

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

∠ADM = ∠CDM (Each 90º)

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, it can be said that

CM = AM = AB