Question

In a triangle DEF G is a point on EF such that EG:GF = 1:2 and H is a point on DG and DH:HG is 3:1 than what is DI:IF where I is a point on DF and E is joind to I which passes through H​

Answer 1

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ABCD is a parallelogram, E and F are the mid-points of AB and CD respectively. GH is any line intersecting AD, EF and BC at G,P and H respectively. Prove that GP=PH.

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Answer

Since E and F are mid-points AB and CD respectively.

∴AE=BE=

2

1

AB and CF=DF=

2

1

CD

But, AB=CD

∴

2

1

AB=

2

1

CD⇒BE=CF

Also, BE∥CF [∵AB∥CD]

∴ BEFC is a parallelogram.

⇒BC∥EF and BE=PH ...(i)

Now, BC∥EF

⇒AD∥EF [∵BC∥AD as ABCD is a ∥

gm

]

⇒AEFD is a parallelogram

⇒AE=GP ...(ii)

But, E is the mid-point of AB.

∴AE=BE

⇒GP=PH [Using (i) and (ii)]