Question

The following figure shows a triangle $ABC$ in which $AB=AC$, $M$ is a point on $AB$ and $N$ is a point on $AC$ such that $BM=CN$. Prove that :
i) $AM=AN$
ii) $\Delta AMC\cong \Delta ANB$
iii) $BN=CM$
iv) $\Delta BMC\cong \Delta CNB$

Answer 1

i)

Since,

AB=AC &

BM=CN

$\Rightarrow AM=AN$ [SS THEOREM]

ii)

Since,

AB=AC &

AM=AN

$\angle A=\angle A$

$\Rightarrow △AMC\approx △ANB$ [SAS THEOREM]

iii)

Since,

AB=AC

AM=AN

BM=CN

$\Rightarrow BN=CM$ [SSS THEOREM]

iv)

Since,

BM=CN

MC=BN

BC=BC

$\Rightarrow △BMC\approx △CNB$ [SSS THEOREM]