Easy
In the given figure,$C$ is the midpoint of $AB$ and $D$ is the midpoint of $AC$. Prove that $AD = \frac{1}{4} AB$. Mention the Euclid's axioms used in it.
(N/A) Given: $C$ is the midpoint of $AB$,so $AC = CB$. Since $AB = AC + CB$,we have $AB = AC + AC = 2AC$. Thus,$AC = \frac{1}{2} AB$.
Also,$D$ is the midpoint of $AC$,so $AD = DC$. Since $AC = AD + DC$,we have $AC = AD + AD = 2AD$. Thus,$AD = \frac{1}{2} AC$.
Substituting $AC = \frac{1}{2} AB$ into the equation $AD = \frac{1}{2} AC$,we get $AD = \frac{1}{2} (\frac{1}{2} AB) = \frac{1}{4} AB$.
Euclid's Axioms used:
$1$. Axiom $(7)$: Things which are half of the same things are equal to one another.
$2$. Axiom $(2)$: If equals are added to equals,the wholes are equal.
Also,$D$ is the midpoint of $AC$,so $AD = DC$. Since $AC = AD + DC$,we have $AC = AD + AD = 2AD$. Thus,$AD = \frac{1}{2} AC$.
Substituting $AC = \frac{1}{2} AB$ into the equation $AD = \frac{1}{2} AC$,we get $AD = \frac{1}{2} (\frac{1}{2} AB) = \frac{1}{4} AB$.
Euclid's Axioms used:
$1$. Axiom $(7)$: Things which are half of the same things are equal to one another.
$2$. Axiom $(2)$: If equals are added to equals,the wholes are equal.
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