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Question
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
- D is the mid-point of AC
- MD ⊥ AC
- CM = MA = `1/2AB`
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Solution
(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 = 180° ...(Co-interior angles)
∠MDC + 90° = 180°
∠MDC = 90°
∴ 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 = `1/2AB` ...(M is the mid-point of AB)
Therefore, it can be said that
CM = AM = `1/2AB`
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