1.

Let `A B C`be an isosceles triangle with `A B=A C`and let `D , E ,F`be the mid-points of `B C ,C A`and `A B`respectively. Show that `A D_|_F E`and `A D`is bisected by `F Edot`

Answer» Let AD intersect FE at M. Join DE and DF.
Now, D and E being the midpoints of the sides BC and CA respectively, we have
DE||AB and `DE=(1)/(2) AB " " ` (by midpoint theorem).
Similarly, DF||AC and `DF = (1)/(2) AC. `
`therefore AB = AC rArr (1)/(2) AB =(1)/(2) AC rArr DE=DF. " " `...(i)
Now, DE||FA and `DE = FA " " [ because "DE||AB and "DE=(1)/(2) AB =FA]`
`rArr " DEAF is a ||gm " rArr "DEAF is a rhombus " [ because DE=DF " from (i), " DE = FA and DF = EA].`
But, the diagonals of a rhombus bisect each other at right angles.
`therefore AD bot FE and AM= MD.`
Hence, ` AD bot FE ` and AD is bisected by FE.


Discussion

No Comment Found

Related InterviewSolutions