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 = 1/2 AB

(i) D is the mid-point of AC

 Considering ∆ABC

We observe that M is the mid point of side AB and DM || BC [Given]

This implies, DC= AD ……….. (NCERT Theorem 8.10)

Hence, D is the mid-point of AC.

(ii) MD ⊥ AC

We know that MD || BC and AC is transversal

This implies, ∠ACB = ∠ADM = 90°

Hence, MD ⊥AC

(iii) CM = MA = ½ AB

Considering ∆ADM and ∆CDM

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

∠CDM = ∠ADM (proved, MD ⊥AC)

DM = DM (common)

∆ADM ≅ ∆CDM (By SAS congruency)

CM = AM (By C.P.C.T.)

CM = AM = ½ AB (M is the mid point of AB)

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