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(ii) MD ⊥ AC(i) D is the mid-point of AC(iii) CM = MA = AB

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(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...
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(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 read less
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