In the adjoining figure, if ABN = ACM, show that AMN ABC. - Vidyadip

Question

In the adjoining figure, if $\Delta$ABN = $\Delta$ACM, show that $\Delta$AMN $\sim$ $\Delta$ABC.

✓

Answer

Triangles $\triangle A B N$ and $\triangle A C M$ lie on the same base line AC.
Given that their areas are equal, triangles on the same base and between the same parallels have equal altitudes.
Hence, the perpendicular distances of points B and M from line AC are equal.
Therefore, segment $B M \| A C$.
Now, in triangle $\triangle A B C$, a line through point M is drawn parallel to side AC, meeting side AC at point N.
By the Basic Proportionality Theorem, the line parallel to one side of a triangle divides the other two sides proportionally.
Thus,
$\frac{A M}{A B}=\frac{A N}{A C}$
Therefore, the corresponding sides of triangles $\triangle A M N$
$\triangle A B Care proportional.$
Hence, by the AA similarity criterion,

Thus, $\triangle A M N \sim \triangle A B C$
Hence, the given statement is proved.

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