Question

IN triangle ABC points M and N on side AB and AC respectively are taken, so that AM=1/4 AB and AN=1/4AC.Prove that MN =1/4BC.

Answer 1

To prove-MN=1/4BC
Construction-Let P and Q be the mid points of AB and AC respectively.Join PQ.
Proof-In triangle APQ,
M and N are the mid points of AP and AQ respectively.
Therefore,
MN=1/2PQ(Using mid point theorum).... (1)
In triangle ABC,
P and Q are the mid points of AB and AC respectively.
Therefore,
PQ=1/2BC (Using mid-point theorum)
Putting PQ=1/2BC in (1),
MN=1/2 (1/2BC)
=>MN=1/4BC.
Hence,Proved.
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Answer 2

$\huge{\bf{\underline{\green{Answer :}}}}$

Given,

• In triangle ABC , point M and N on sides AB and AC are taken so that AM = $\bold{\frac{1}{4}AB}$  and AN = $\bold{\frac{1}{4}AC}$

Prove :

$\implies \bold{MN = \frac{1}{4}BC}$

Construction :  

⇒Join EF where E and F are the mid points of AB and AC .

Proof :

Mid Point Theorem :

⇒A line segment joining the mid - points of any two sides of a triangle is parallel to the third side and is half of it .

∵E is the mid point of AB and F is the mid point of AC

∴EF||BC So , EF = $\bold{\frac{1}{2}BC}$.....................[ 1) equation ]

Now , AE = $\bold{\frac{1}{2}AB}$...............[ by construction ]

AM = $\bold{\frac{1}{4}AB}$.....................[ given]

∴ AM = $\bold{\frac{1}{2}AE}$

Similarly , AN = $\bold{\frac{1}{2}AF}$

⇒M and N are the mid point of AE and AF .

∴ MN||EF And MN = $\bold{\frac{1}{2}EF}$.............[By mid point theorem}

Now ,

$\implies \bold{MN = \bold{\frac{1}{2}EF}}$

So, MN  $\implies \bold{\frac{1}{2} (\bold{\frac{1}{2}BC}})$...................[ From 1 equation ]

Therefore ,

MN=  $\bold{\frac{1}{4}BC}$....................Hence proved