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.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(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.

Explore More

Similar Questions

The Greeks emphasized on:

Solve the following question using appropriate Euclid's axiom:
In the figure:
$BM = BN$,$M$ is the mid-point of $AB$ and $N$ is the mid-point of $BC$. Show that $AB = BC$.

Which of the following needs a proof?

Consider two 'postulates' given below:
$(i)$ Given any two distinct points $A$ and $B$,there exists a third point $C$ which is in between $A$ and $B$.
$(ii)$ There exist at least three points that are not on the same line.
Do these postulates contain any undefined terms? Are these postulates consistent? Do they follow from Euclid's postulates? Explain.

Euclid's fifth postulate is:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo