HIII.....
GOOD MORNING
EVERYONE
PLEASE SOLVE QUESTION NO 36 IN THE ATTACHMENT AND ANSWER IS OPTION D .
THANKS......☺☺☺☺
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HIII.....
GOOD MORNING
EVERYONE
PLEASE SOLVE QUESTION NO 36 IN THE ATTACHMENT AND ANSWER IS OPTION D .
THANKS......☺☺☺☺
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Hey purro
Here is ur answer
Given: AD is the median of ΔABC. E is the mid point of AD. BE produced meets AD at F
To Prove :3/4AC
Construction: From Point D, draw DGBF.
Proof: In ΔADG, E is the mid-point of AD and EF||DG.
∴F is the mid point of AG [Converse of the mid point theorem]
⇒ AF = FG ... (i)
In ΔBCF, D is the mid point of BC and DG||BF
∴ G is the mid point of CF
⇒ FG = GC ... (ii)
From (i) and (ii), we get,
AF = FG = GC ... (iii)
Now, AF + FG + GC = AC
⇒ AF + AF + AF = AC [Using (iii)]
⇒ 3AF = AC
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