Question 15 Marks

In the:
$i. AB = BC, M$ is the mid$-$point of $AB$ and $N$ is the mid$-$ point of $BC.$ Show that $AM = NC.$
$ii. BM = BN, M$ is the mid$-$point of $AB$ and $N$ is the mid$-$point of $BC.$ Show that $AB = BC.$

Answer

Given, $AB = BC ....(i) M$ is the mid$-$point of $AB.$
$\therefore\ \text{AM}=\text{MB}=\frac{1}{2}\text{AB}\ ...(\text{ii})$ And $N$ is the mid$-$point $BC.$
$\therefore\ \text{BN}=\text{NC}=\frac{1}{2}\text{BC}\ ...(\text{iii})$
According to euclid's axiom, things which are halves of the same things are equal to one another.
From Eq. $(i) AB = BC$ On multiplying both sides by $\frac{1}{2},$
we get $\frac{1}{2}\text{AB}=\frac{1}{2}\text{BC}$
$\Rightarrow\text{AM}=\text{NC} [$Using Eqs. $(ii)$ and $(iii)]$
Given, $BM = BN ....(i) M$ is the mid$-$point of $AB$
$\therefore\ \text{AM}=\text{BM}=\frac{1}{2}\text{AB}$
$\Rightarrow\ 2\text{AM}=2\text{BM}=\text{AB}\ ....(\text{ii})$ and $N$ is the mid$-$point of $BC.$
$\therefore\ 2\text{BN}=2\text{NC}=\text{BC}\ ...(\text{iii})$
According to Euclid's axiom, things which are double of the same thing are equal to one another.
On multiplying both sides of Eq. $(i)$ by $2,$ we get
$\Rightarrow\ 2\text{BM}=2\text{BN}$
$\text{AB}=\text{BC} [$Using Eqs. $(ii)$ and $(iii)]$

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