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14/07/2021 12:13 pm
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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
1 Answer
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14/07/2021 12:18 pm
(i) In ΔACB,
M is the midpoint of AB and MD || BC
D is the midpoint of AC (Converse of mid point theorem)
(ii) ∠ACB = ∠ADM (Corresponding angles)
also, ∠ACB = 90°
∠ADM = 90° and MD ⊥ AC
(iii) In ΔAMD and ΔCMD,
AD = CD (D is the midpoint of side AC)
∠ADM = ∠CDM (Each 90°)
DM = DM (common)
ΔAMD ≅ ΔCMD [SAS congruency]
AM = CM [CPCT]
AM = 1/2 AB (M is midpoint of AB)
Hence, CM = MA = 1/2 AB