Ceva’s Theorem MCQ Quiz - Objective Question with Answer for Ceva’s Theorem - Download Free PDF

Concept:

Ceva's Theorem

According to this theorem, if AD, BE, CF are concurrent lines meeting at the point O of a Δ ABC then

$\frac{AF}{FB}\times \frac{BD}{DC}\times \frac{CE}{EA}=1$

Calculation:

From Ceva's Theorem, we can write as

$\frac{AF}{FB}\times \frac{BD}{DC}\times \frac{CE}{EA}=1$

⇒ AF × BD × CE = FB × DC × EA

On comparing the above with the given

BD × CE × X = DC × EA × Y

The answer to the question is AF and FB

Concept:

Ceva's Theorem

According to this theorem, if AD, BE, CF are concurrent lines meeting at the point O of a Δ ABC then

$\frac{AF}{FB}\times \frac{BD}{DC}\times \frac{CE}{EA}=1$

Given:

In ΔABC

D, E, F be the points on lines BC, CA, and AB respectively such that lines AD, BE, and CF are concurrent.

${\displaystyle \frac{AF}{FB}}\times {\displaystyle \frac{BD}{DC}}={\displaystyle \frac{1}{3}}$

Calculation:

In ΔABC

Using the result,

$\frac{AF}{FB}\times \frac{BD}{DC}\times \frac{EC}{EA}=1$

$\frac{1}{3}\times \frac{EC}{EA}=1$

$\frac{EA}{EC}=\frac{1}{3}$

Concept:

Ceva's Theorem

According to this theorem, if AD, BE, CF are concurrent lines meeting at the point O of a Δ ABC then

$\frac{AF}{FB}\times \frac{BD}{DC}\times \frac{CE}{EA}=1$

Given:

In ΔABC

D, E, F be the points on lines BC, CA, and AB respectively such that lines AD, BE, and CF are concurrent.

${\displaystyle \frac{\mathrm{C}\mathrm{E}}{\mathrm{E}\mathrm{A}}}={\displaystyle \frac{3}{5}},{\displaystyle \frac{\mathrm{B}\mathrm{D}}{\mathrm{D}\mathrm{C}}}={\displaystyle \frac{5}{7}}$

AB = 14 cm

Calculation:
Using the result as

$\frac{AF}{FB}\times \frac{BD}{DC}\times \frac{CE}{EA}=1$

⇒ $\frac{AF}{FB}\times \frac{5}{7}\times \frac{3}{5}=1$

⇒ $\frac{AF}{FB}=\frac{7}{3}$

AB = 14 cm (given)

10 units = 14 ( Since, AF = 7 units and FB = 3 units, on adding both we get AB as 10 units )

1 unit = 1.4

AF = 7 × 1.4 = 9.8 cm

FB = 3 × 1.4 = 4.2 cm

Concept:

Menelaus' theorem: Let ABC be a triangle and D, E and F be points on the line formed from AD, BC and AB respectively. If D, E and F are collinear then $\frac{AF}{FB}\times \frac{BE}{EC}\times \frac{CD}{AD}=1$ 

Ceva's Theorem: Let ABC is a triangle and AD, BE and CF are Cevians or intersecting at a common point O then $\frac{AE}{EC}\times \frac{CD}{BD}\times \frac{BF}{AF}=1$

Ceva's and Menelaus's Theorem are dual of each other.

Calculation:

1.If the quadrilateral is cyclic, then the product of two diagonals is equal to the sum of the products of the opposite side lengths (Cyclic Quadrilateral property) - Correct

2. Ceva's theorem is concerned with concurrency of the lines. - Correct

3. Menelau's theorem is concerned with the collinearity of the points - Correct

4. Menelau's theorem is a not dual version of Ceva's theorem.- False

Concept:

Ceva's Theorem

According to this theorem, if AD, BE, CF are concurrent lines meeting at the point O of a Δ ABC then

$\frac{AF}{FB}\times \frac{BD}{DC}\times \frac{CE}{EA}=1$

Given:

In ΔABC

D, E, F be the points on lines BC, CA, and AB respectively such that lines AD, BE, and CF are concurrent.

${\displaystyle \frac{AF}{FB}}={\displaystyle \frac{1}{2}},{\displaystyle \frac{BD}{DC}}={\displaystyle \frac{2}{3}}$

Calculation:

In ΔABC

Using the result 

$\frac{AF}{FB}\times \frac{BD}{DC}\times \frac{EC}{EA}=1$

$\frac{1}{2}\times \frac{2}{3}\times \frac{EC}{EA}=1$

$\frac{EA}{EC}=\frac{1}{3}$

Concept:

Ceva's Theorem

According to this theorem, if AD, BE, CF are concurrent lines meeting at the point O of a Δ ABC then

$\frac{AF}{FB}\times \frac{BD}{DC}\times \frac{CE}{EA}=1$

Given:

In ΔABC

D, E, F be the points on lines BC, CA, and AB respectively such that lines AD, BE, and CF are concurrent.

${\displaystyle \frac{AF}{FB}}\times {\displaystyle \frac{BD}{DC}}={\displaystyle \frac{1}{3}}$

Calculation:

In ΔABC

Using the result,

$\frac{AF}{FB}\times \frac{BD}{DC}\times \frac{EC}{EA}=1$

$\frac{1}{3}\times \frac{EC}{EA}=1$

$\frac{EA}{EC}=\frac{1}{3}$

Concept:

Menelaus' theorem: Let ABC be a triangle and D, E and F be points on the line formed from AD, BC and AB respectively. If D, E and F are collinear then $\frac{AF}{FB}\times \frac{BE}{EC}\times \frac{CD}{AD}=1$ 

Ceva's Theorem: Let ABC is a triangle and AD, BE and CF are Cevians or intersecting at a common point O then $\frac{AE}{EC}\times \frac{CD}{BD}\times \frac{BF}{AF}=1$

Ceva's and Menelaus's Theorem are dual of each other.

Calculation:

1.If the quadrilateral is cyclic, then the product of two diagonals is equal to the sum of the products of the opposite side lengths (Cyclic Quadrilateral property) - Correct

2. Ceva's theorem is concerned with concurrency of the lines. - Correct

3. Menelau's theorem is concerned with the collinearity of the points - Correct

4. Menelau's theorem is a not dual version of Ceva's theorem.- False

Concept:

Ceva's Theorem

According to this theorem, if AD, BE, CF are concurrent lines meeting at the point O of a Δ ABC then

$\frac{AF}{FB}\times \frac{BD}{DC}\times \frac{CE}{EA}=1$

Given:

In ΔABC

D, E, F be the points on lines BC, CA, and AB respectively such that lines AD, BE, and CF are concurrent.

${\displaystyle \frac{AF}{FB}}={\displaystyle \frac{1}{2}},{\displaystyle \frac{BD}{DC}}={\displaystyle \frac{2}{3}}$

Calculation:

In ΔABC

Using the result 

$\frac{AF}{FB}\times \frac{BD}{DC}\times \frac{EC}{EA}=1$

$\frac{1}{2}\times \frac{2}{3}\times \frac{EC}{EA}=1$

$\frac{EA}{EC}=\frac{1}{3}$

Concept:

Ceva's Theorem

According to this theorem, if AD, BE, CF are concurrent lines meeting at the point O of a Δ ABC then

$\frac{AF}{FB}\times \frac{BD}{DC}\times \frac{CE}{EA}=1$

Given:

In ΔABC

D, E, F be the points on lines BC, CA, and AB respectively such that lines AD, BE, and CF are concurrent.

$\frac{\mathrm{C}\mathrm{E}}{\mathrm{E}\mathrm{A}}=\frac{5}{6},\frac{\mathrm{B}\mathrm{D}}{\mathrm{D}\mathrm{C}}=\frac{4}{3}$

AB = 5.7 cm

Calculation:
Using the result as

$\frac{AF}{FB}\times \frac{BD}{DC}\times \frac{CE}{EA}=1$

⇒ $\frac{AF}{FB}\times \frac{4}{3}\times \frac{5}{6}=1$

⇒ $\frac{AF}{FB}=\frac{18}{20}=\frac{9}{10}$

Concept:

Ceva's Theorem

According to this theorem, if AD, BE, CF are concurrent lines meeting at the point O of a Δ ABC then

$\frac{AF}{FB}\times \frac{BD}{DC}\times \frac{CE}{EA}=1$

Given:

In ΔABC

D, E, F be the points on lines BC, CA, and AB respectively such that lines AD, BE, and CF are concurrent.

${\displaystyle \frac{\mathrm{C}\mathrm{E}}{\mathrm{E}\mathrm{A}}}={\displaystyle \frac{3}{5}},{\displaystyle \frac{\mathrm{B}\mathrm{D}}{\mathrm{D}\mathrm{C}}}={\displaystyle \frac{5}{7}}$

AB = 14 cm

Calculation:
Using the result as

$\frac{AF}{FB}\times \frac{BD}{DC}\times \frac{CE}{EA}=1$

⇒ $\frac{AF}{FB}\times \frac{5}{7}\times \frac{3}{5}=1$

⇒ $\frac{AF}{FB}=\frac{7}{3}$

AB = 14 cm (given)

10 units = 14 ( Since, AF = 7 units and FB = 3 units, on adding both we get AB as 10 units )

1 unit = 1.4

AF = 7 × 1.4 = 9.8 cm

FB = 3 × 1.4 = 4.2 cm

Concept:

Ceva's Theorem

According to this theorem, if AD, BE, CF are concurrent lines meeting at the point O of a Δ ABC then

$\frac{AF}{FB}\times \frac{BD}{DC}\times \frac{CE}{EA}=1$

Given:

In ΔABC

D, E, F be the points on lines BC, CA, and AB respectively such that lines AD, BE, and CF are concurrent.

${\displaystyle \frac{\mathrm{C}\mathrm{E}}{\mathrm{E}\mathrm{A}}}={\displaystyle \frac{2}{7}},{\displaystyle \frac{\mathrm{B}\mathrm{D}}{\mathrm{D}\mathrm{C}}}={\displaystyle \frac{9}{4}}$

AB = 4.6 cm

Calculation:
Using the result as

$\frac{AF}{FB}\times \frac{BD}{DC}\times \frac{CE}{EA}=1$

⇒ $\frac{AF}{FB}\times \frac{2}{7}\times \frac{9}{4}=1$

⇒ $\frac{AF}{FB}=\frac{14}{9}$

AB = 4.6 cm (given)

23 units = 4.6 ( Since, AF = 14 units and FB = 9 units, on adding both we get AB as 23 units)

1 unit = 0.2

AF = 14 × 0.2 = 2.8 cm

FB = 9 × 0.2 = 1.8 cm

Concept:

Ceva's Theorem

According to this theorem, if AD, BE, CF are concurrent lines meeting at the point O of a Δ ABC then

$\frac{AF}{FB}\times \frac{BD}{DC}\times \frac{CE}{EA}=1$

Calculation:

From Ceva's Theorem, we can write as

$\frac{AF}{FB}\times \frac{BD}{DC}\times \frac{CE}{EA}=1$

⇒ AF × BD × CE = FB × DC × EA

On comparing the above with the given

BD × CE × X = DC × EA × Y

The answer to the question is AF and FB

Concept:

Ceva's Theorem

According to this theorem, if AD, BE, CF are concurrent lines meeting at the point O of a Δ ABC then

$\frac{AF}{FB}\times \frac{BD}{DC}\times \frac{CE}{EA}=1$

Given:

In ΔABC

D, E, F be the points on lines BC, CA, and AB respectively such that lines AD, BE, and CF are concurrent.

${\displaystyle \frac{AF}{FB}}\times {\displaystyle \frac{BD}{DC}}={\displaystyle \frac{1}{3}}$

Calculation:

In ΔABC

Using the result,

$\frac{AF}{FB}\times \frac{BD}{DC}\times \frac{EC}{EA}=1$

$\frac{1}{3}\times \frac{EC}{EA}=1$

$\frac{EA}{EC}=\frac{1}{3}$

Concept:

Menelaus' theorem: Let ABC be a triangle and D, E and F be points on the line formed from AD, BC and AB respectively. If D, E and F are collinear then $\frac{AF}{FB}\times \frac{BE}{EC}\times \frac{CD}{AD}=1$ 

Ceva's Theorem: Let ABC is a triangle and AD, BE and CF are Cevians or intersecting at a common point O then $\frac{AE}{EC}\times \frac{CD}{BD}\times \frac{BF}{AF}=1$

Ceva's and Menelaus's Theorem are dual of each other.

Calculation:

1.If the quadrilateral is cyclic, then the product of two diagonals is equal to the sum of the products of the opposite side lengths (Cyclic Quadrilateral property) - Correct

2. Ceva's theorem is concerned with concurrency of the lines. - Correct

3. Menelau's theorem is concerned with the collinearity of the points - Correct

4. Menelau's theorem is a not dual version of Ceva's theorem.- False

Concept:

Ceva's Theorem

According to this theorem, if AD, BE, CF are concurrent lines meeting at the point O of a Δ ABC then

$\frac{AF}{FB}\times \frac{BD}{DC}\times \frac{CE}{EA}=1$

Given:

In ΔABC

D, E, F be the points on lines BC, CA, and AB respectively such that lines AD, BE, and CF are concurrent.

${\displaystyle \frac{AF}{FB}}={\displaystyle \frac{1}{2}},{\displaystyle \frac{BD}{DC}}={\displaystyle \frac{2}{3}}$

Calculation:

In ΔABC

Using the result 

$\frac{AF}{FB}\times \frac{BD}{DC}\times \frac{EC}{EA}=1$

$\frac{1}{2}\times \frac{2}{3}\times \frac{EC}{EA}=1$

$\frac{EA}{EC}=\frac{1}{3}$