Question

M
5. In the figure, triangle ABC ~ triangle MNO,
D is the midpoint of side AC and P
is the midpoint of side MO.
Prove : (i) triangle ABD ~ triangleMNP
(ii) BD/NP=
AB/MN
​

Answer 1

Answer:

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Answer 2

5. In the figure, triangle ABC ~ triangle MNO,  D is the midpoint of side AC and P  is the midpoint of side MO.

Given,

Δ ABC ~ Δ MNO

⇒ AB = MN  

    BC = NO

    AC = MO  

D is the midpoint of side AC and P  is the midpoint of side MO

⇒ AC = AD + DC

⇒ MO = MP + PO

Now consider,

In Δ ABD and Δ MNP

AB = MN       (sides of similar triangles are similar)

∠ A = ∠ M      (∵ Δ ABC ~ Δ MNO )

BD = NP        (BD ⊥ AC  and NP ⊥ MO)

∴ Δ ABD ~ Δ MNP .................(i)

using (i) we have,

BD/AB = NP/MN   ( c.p.c.t )

∴ BD/NP = AB/MN ..................(ii)

Hence proved.