In □m ABCD, M is the mid-point of 
. 
and 
intersect in N. Prove that DN = 2MN.
In 
ABCD, M is the mid-point of
. 
and 
intersect in N.
To prove : DN = 2MN
Proof: M is the mid-point of BC
MB = MC
…(1)
In ∆MBN and ∆MCD
∠BMN ≅ ∠CMD (Vertically opposite angles)
∠MNB ≅ ∠MDC (Alternate angles)
The correspondence MBN↔MCD is a similarity
(from 1)
MN = MD
Now, D-M-N
DN = DM + MN
DN = DM + MN
DN = MN + MN (MD = MN)
DN = 2MN
Hence proved.
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is an altitude to hypotenuse. AB = 8, BC = 6. Find the area of ΔBDC.
and 
intersects 
in M and 
in O. Prove that AM2 = MT • MO.
and 
in ΔABC. If the area of ΔAPQ = 12√3, find the area of ΔABC.
= {0}. Prove that the area of ΔOAB = 
(area of