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Ceva's Theorem is based on the triangular plain geometry. It states that the ratio of the products of the pairs of segments of a triangle is equal to the value of one. This is an important theorem of the triangle that lies on the Euclidean plane geometry. The mathematical theorem deals with the rules of modern mathematics. This theorem comes under the affine geometry concept. However, in this theorem, the collinear property has been used. This theorem is mostly used in solving affine plane geometry and in proving the cevians congruence in the triangle.
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Keyterms: Geometry, Triangle, Congruence in the triangles, Angle, Area of Triangle
Read more: Pythagoras Theorem
What is Ceva's Theorem?
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The Ceva's Theorem was first discovered by an Italian scientist named Tommaso Ceva in the 18th century. He states that at any given triangle ABC, the segments drawn from the vertex of the triangle to their opposite sides of a triangle are precisely concurrent when the ratio of the products of all the drawn segments of the triangle from each side is equal to one.
Ceva’s Theorem
This theorem comes under the affine geometry concept which means that it can be stated and proven without using the concepts of lengths, angles, and areas. However, in this theorem, the collinear property has been used. This theorem is mostly used in solving affine plane geometry and in proving the cevians congruence in the triangle.
Ceva's Theorem Statement
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Consider a triangle ABC having a point O lies internally inside the triangle. Let the lines AO, BO, and CO touch the sides of a triangle BC, CA, and AB at the points D, E, and F respectively.
Ceva’s Theorem
According to the Ceva's Theorem
AF / FB × BD / DC × CE / EA = 1
And the converse of Ceva's Theorem states that the points of contact of the triangle D, E, and F on the sides BC, CA, and AB respectively then the lines which are AD, BE, and CF are concurrent at the points O.
AF / FB × BD / DC × CE / EA = 1
Proof of Ceva's Theorem
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The Ceva's Theorem can be explained as per the following steps given below
Step 1: Draw a triangle ABC with a point P inside the triangle and draw lines AP, BP, and CP on it to extend it to their intersections with the opposite sides in points D, E, and F of the triangle.
Step 2: With the help of a ruler, draw a line that passes through B and perpendicular to the side AC. Name this point as Z.
Ceva’s theorem proof
Step 3: Now, have a look at both the areas of two triangles. Triangle ABE & CBE both have the same height, BZ. The triangle ABE & CBE has a base AE and CE. Therefore,
ΔABE : ΔCBE = (AE / CE)
Step 4: Now, let the smaller triangle APE and CPE as given above. Therefore it can be written as,
ΔAPE : ΔCPE = (AE / CE)
Step 5: Now If we look at the triangles APB and CPB and on considering there areas we get as,
Area of triangle APB = Area of triangle ABE - Area of triangle APE and
Area of triangle BPC = Area of triangle CBE - Area of triangle CPE
We proved that,
ΔABE :ΔCBE = (AE / CE) and ΔAPE :ΔCPE = (AE / CE)
So, therefore
ΔABP :ΔCBP = ΔABP /ΔBPC = AE / CE
Hence the same is true for the triangle CPA and CPB and the triangles APB and APC.
Therefore
ΔCPB : ΔCPA = ΔCPB /ΔCPA = BF/AF
and ΔAPC : ΔAPB = ΔAPC /ΔAPB = CD/BD
Now if we multiply it together then we get,
ΔABP /ΔBPC × ΔCPB /ΔCPA × ΔAPC / ΔAPB = AE / CE × BF/AF × CD/BD
After solving the above equation we get
AE / CE × BF / AF × CD / BD = 1
Hence we proved Ceva's Theorem.
Read more: Nature of roots of quadratic equations
Converse of Ceva's Theorem
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As we know,
AE / CE × BF/AF × CD/BD = 1
Take a triangle ABC having a point P inside the triangle and segments FP and DP.
If we show that the line BPE is collinear then the converse of Ceva's Theorem is also proved.
The converse of Ceva's Theorem can be proved as,
From Ceva's Theorem, we can write
ΔABP :ΔCBP = ΔABP /ΔBPC = AE / CE
and ΔCPB : ΔCPA = ΔCPB /ΔCPA = BF/ AF
If we multiply the above equation we get,
ΔABP /ΔBPC × ΔCPB /ΔCPA = AE / CE × BF/AF
After simplifying we get,
ΔABP /ΔCBP × ΔCPB /ΔCPA = AE / CE × BF/AF
As we know,
AE / CE × BF/AF × CD/BD = 1
AE / CE × BF/AF = BD / CD
Therefore we can say that
ΔAPC /ΔAPB = BD/CD = AE / CE × BF/AF
Now if we extend the other segment BP to meet the side AC at point Z,
ΔAPC /ΔAPB = BD/CD
Therefore we can say that D = Z and the proof is complete.
Hence if we extend the segments to the point P, it will intersect at D. This proved the converse of Ceva's Theorem. Thus, the medians of the triangles are concurrent.
Things to Remember
- Ceva’s theorem is for the affine Euclidean plane geometry in which the vertices of the triangle or cevians of the triangle form a concurrent point on the triangle.
- The lines which pass through a common point and intersect both the vertices as well as the opposite side of the triangle corresponding to the vertex is known as Cevian.
- Angles and medians bisectors of triangles are some special cases of the Cevians.
Sample Questions
Ques. What does Ceva's theorem state? (2 marks)
Ans. Ceva's Theorem expresses that in a triangle ABC, the lines from A, B, and C to the opposite sides of the triangle are simultaneous exactly when the product of the proportions of the sets of the pairs of segments on each side of the triangle is equivalent to 1.
Ques. In a triangle ABC, if DE || BC, AE = 8 cm, EC = 2 cm and BC = 6 cm, then find the value of DE? (2 marks)
Ans. Draw a triangle ABC in which DE is parallel to BC.
In ΔADE and ΔABC,
∠DAE = ∠BAC (Common)
∠ADE – ∠ABC (Corresponding angles)
ΔADE – ΔΑΒC (AA corollary)
Hence we can say
AE / AC = DE / BC
8 / 8+2 = DE / 6
10 DE = 48
DE = 4.8 cm
Ques. What is the Angle bisector theorem in Triangle? (2 marks)
Ans. The angles which bisect the opposite sides of the triangle in such a way that the ratio of the two line segments are proportional with the ratios of the other two sides of the triangle is known as the angle bisector theorem.
Ques. What do you mean by the AAA similarity theorem? (2 marks)
Ans. The angles similarity theorem can be formulated as the two triangles having their corresponding angles are equal if and only if the corresponding sides of the triangle are equal.
Ques. How do you prove the converse of Ceva's theorem? (2 marks)
Ans. The converse of Ceva's Theorem expresses that if the product of the proportions of the three sides of a triangle when divided by three results and is equivalent to 1, then, at that point, it implies that the lines that join these three points to the contrary vertices of the triangle are concurrent.
Ques. Are cevians lines concurrent? (2 marks)
Ans. Cevians come in sets of three, similar to the three medians of a triangle, the three angle bisectors, or the three heights. In every one of these cases, the three cevians are concurrent: The medians meet at the centroid, the point bisectors meet at the incenter, and the altitudes meet at the orthocenter of a triangle.
Ques. What is the formula for calculating the length of a cevian? (2 marks)
Ans. The length of the cevian can be calculated by the following three formulas given below
- If the cevian is an altitude of a triangle, its length is given by the formula: d2= b2 − n2 = c2 − m2.
- If the cevian is a median of a triangle, its length is given by the formula: m(b2+c2) = a(d2+m2).
- If the cevian is a bisector of a triangle, its length is given by the formula: (b+c)2 = a2(d2mn+1).
Ques. If a triangle ABC is congruent to the triangle RPQ, and AB = 3 cm BC = 5 cm, AC = 6 cm, RP = 6 cm and PQ = 10, then find the value of QR? (2 marks)
Ans. It is given that the ΔABC ~ ΔRPQ
Then AB / RP = BC / PQ = AC / RQ
Putting all the given values we get
3 / 6 = 5 / 10 = 6 / QR
After solving the above equation we get
1 / 2 = 6 / QR
Hence, QR = 12 cm.
Ques. In ΔABC, DE || BC, find the value of x. (CBSE 2015) (2 marks)
Ans: In ΔABC, DE || BC …[Given]
AD/BD = AE/EC
\(\frac{x}{x + 1} = \frac{x + 3}{x + 5}\) (by Thales Theorem)
x(x + 5) = (x + 3)(x + 1)
x2 + 5x = x2 + 3x + x + 3
x2 + 5x – x2 – 3x – x = 3
∴ x = 3 cm
Ques. In the given figure, if DE || BC, AE = 8 cm, EC = 2 cm and BC = 6 cm, then find DE. (CBSE 2014) (2 marks)
Ans. In ΔADE and ΔABC,
∠DAE = ∠BAC …[Common]
∠ADE – ∠ABC … [Corresponding angles]
ΔADE – ΔΑΒC …[AA corollary]
AE/AC = DE/BC
\(\frac{8}{8 + 2} = \frac{DE}{6}\) (Corresponding sides are proportional)
10DE = 48
DE = 4.8
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