Question

In the given figure, 2AD = BD. E is mid-point of BD and F is mid-point of AC and EC ll BH. Prove that:
(1) DF Il BH
(2) AH = 3 AF​

Answer 1

To prove that :

(1) DF ║ BH

(2) AH = 3 AF

Given :

"E" is the mid point of BD.

"F" is the mid point of AC.

2 AD = BD  and  EC ║ BH

(1) DF ║ BH :

From the figure,

"E" is the mid point of BD.

2 DE = BD   -----> Equation (1)

2 AD = BD   -----> Equation (2)

From the equation (1) and (2), we get

2 DE = 2 AD  

That is, DE = AD.

Where, D is the mid point of AE.

            F is the mid point of AC (Given).

So, DF ║ EC    -----> Equation (3)

     EC ║ BH   -----> Equation (4)

From equation (3) and (4), we get

  ∴ DF ║ BH.

(2) AH = 3 AF :

Here, "E" is the mid point of BD and EC ║ BH.

Now, "C" is the mid point of AH.

Here, FC = CH   -----> Equation (5)

"F" is the mid point of AC.

Here, AF = FC   -----> Equation (6)

From equation (5) and (6), we get

FC = AF = CH

(i.e.,) AF = $\frac{1}{3}$ AH

   ∴ AH = 3 AF

Therefore, the condition has been proved.

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