In a $Δ A B C$ , M and N are points on the sides AB and AC respectively such that BM = CN. If $∠ B = ∠ C$ then show that MN || BC.

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Solution

$A B = A C (s i n c e ∠ B = ∠ C)$ ....(1)

$B M = C N (g i v e n)$

from 1 , AM+MB = AN +NC

AM=AN ( MB=NC)

$\frac{A M}{B M}$ = $\frac{A N}{N C}$

since MN is dividing the sides AB and AC in proportion , by Basic proportionality theorem MN || BC

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In a ΔABC, M and N are points on the sides AB and AC respectively such that BM = CN. If ∠B=∠C then show that MN || BC.

AB=AC (since∠B=∠C) ....(1) BM=CN (given) from 1 , AM+MB = AN +NC AM=AN ( MB=NC) AMBM =ANNC since MN is dividing the sides AB and AC in proportion , by Basic proportionality theorem MN || BC

In a $Δ A B C$ , M and N are points on the sides AB and AC respectively such that BM = CN. If $∠ B = ∠ C$ then show that MN || BC.

$A B = A C (s i n c e ∠ B = ∠ C)$ ....(1)

$B M = C N (g i v e n)$

from 1 , AM+MB = AN +NC

AM=AN ( MB=NC)

$\frac{A M}{B M}$ = $\frac{A N}{N C}$

since MN is dividing the sides AB and AC in proportion , by Basic proportionality theorem MN || BC