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=12AB
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Solution
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Given : AM=MB & MD∥BC
Since MD∥BC
∠ADM=∠ACB (Corresponding angles)
∠ADM=90∘
Hence , MD⊥AC
Proved.
Since DM∥BC, By Basic proportionality Theorem
(ADAC)=(AMAB)=12
⇒AC=2AD
Thus D is mid point of AC Proved!
In ΔADM & ΔCDM
AD=CD (proved above)
∠ADM=∠CDM(=90∘)
DM=DM (common side)
∴ΔADM≅ΔCDM by SAS
Hence by cpct
AM=CM
But AM=12AB
∴AM=CM=12(AB)
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