Question 15 Marks
In the:
- AB = BC, M is the mid-point of AB and N is the mid- point of BC. Show that AM = NC.
- BM = BN, M is the mid-point of AB and N is the mid-point of BC. Show that AB = BC.
Answer
View full question & 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)]
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)]