A cevian is a line segment that extends from one vertex of a triangle to the opposite side. In the diagram, AD is a cevian, from A.

**Concurrency Theorems**

A cevian is a line segment that extends from one vertex of a
triangle to the opposite side. In the diagram, AD is a cevian, from *A*.

**Special cevians**

(i) A median is a cevian that divides the opposite side into two congruent(equal) lengths.

(ii) An altitude is a cevian that is perpendicular to the opposite side.

(ii) An angle bisector is a cevian that bisects the corresponding angle.

Let ABC be a triangle and let *D,E,F* be points on lines *BC*,
*CA, AB *respectively. Then the cevians *AD, BE, CF *are
concurrent if and only if = 1 where the lengths are directed. This also
works for the reciprocal of each of the ratios *B *as the reciprocal of 1
is 1.

**Note**

The cevians do not necessarily lie within the triangle, although
they do in the diagram.

A necessary and sufficient condition for points *P*, *Q*,
*R* on the respective sides BC, *CA*, *AB* (or their extension)
of a triangle *ABC *to be collinear is that where all segments
in the formula are directed segments.

**Note**

·
Menelaus theorem can also be given as *BP *×* CQ *×*AR
*= −*PC *×* QA*×*RB.*

·
If BP is replaced by** ***PB*** **(or)** ***CQ***
**by** ***QC*** **(or)** ***AR*** **by** ***RA*,
or if** **any one of the six directed line segments *BP*, *PC*, *CQ*,
*QA*, *AR*,* RB *is interchanged, then the product will be* *1.

**Example 4.32 **Show that in a triangle, the medians are
concurrent.

Medians are line segments joining each vertex to the** **midpoint of the
corresponding opposite sides.

Thus medians are the cevians where *D*, *E*, *F*
are midpoints of *BC*,* CA *and* AB *respectively.

Since *D* is a midpoint of *BC*, *BD* = *DC*
so BD/DC =1 ………..(1)

Since, *E* is a midpoint of *CA*, *CE* = *EA*
so CE/EA =1 ………..(1)

Since, *F* is a midpoint of *AB*, *AF* = *FB*
so AF/FB =1 ………..(1)

Thus, multiplying (1), (2) and (3) we get,

And so, Ceva’s theorem is satisfied.

Hence the Medians are concurrent.

In Fig.4.73,** ***ABC*** **is a triangle with** **∠*B*** **=** **90** **,** ***BC*** **=** **3** **cm and** ***AB*** **=** **4** **cm. *D *is point on* AC *such that* AD *=* *1* *cm
and* E *is the midpoint of *AB*. Join* D *and* E *and extend* DE *to meet*
CB *at* F*. Find* BF*.

Consider** **D

By Menelaus’ theorem

By assumption, *AE* = *EB* = 2, *DA* = 1 and

*FC *=* FB *+* BC *=* BF *+* *3

By Pythagoras theorem, *AC* ^{2} = *AB*^{2}
+ *BC* ^{2} = 16 + 9 = 25. Therefore *AC* = 5

and So, *CD *=* AC *–* AD *= 5* *–* *1
= 4.

Substituting the values of *FC*, *AE*, *EB*, *DA*,
*CD* in (1),

4*BF* = *BF* + 3

4*BF* – *BF* = 3 therefore *BF* = 1

Suppose** ***AB*,** ***AC*** **and** ***BC*** **have lengths** **13, 14** **and** ***AFFB *15 respectively. If *AF/FB*
= 2/5 and *CE/EA* = 5/8. Find *BD* and *DC*.

Given that *AB* = 13 , *AC* = 14 and *BC* = 15.

Let *BD* = *x* and *DC* = *y*

Using Ceva’s theorem, we have,

*BC *=* BD *+* DC *=* *15 so, *x* + *y* =
15

From (2), using *x* = 4*y* in (3) we get, 4*y* + *y* = 15 gives 5*y*
= 15 then *y* = 3

Substitute *y* = 3 in (3) we get, *x* = 12 . Hence *BD* = 12 , *DC*
= 3.

In a garden containing several trees, three particular trees** ***P*,** ***Q*,** ***R*** **are located in the
following way, *BP* = 2 m, *CQ* = 3 m, *RA *=* *10* *m,*
PC *=* *6* *m,* QA *=* *5* *m,* RB *=* *2*
*m, where *A*,* B*,* C *are points such that* P *lies on*
BC*,* Q *lies on *AC *and* R *lies on* AB*. Check
whether the trees P,* Q*,* R *lie on a same straight line.

By Meanlau’s theorem, the trees* **P*,* **Q*,* **R*** **will be collinear (lie
on same straight line)

Given *BP* =2 m, *CQ* =3 m, *RA* =10 m, *PC*
=6 m,

*QA *= 5* *m and* RB *= 2* *m

Substituting these values in (1) we get, *BP/PC* × *CQ/QA* × *RA/RB*
= 2/6 × 3/5 × 10/2 = 60/60 = 1

Hence the trees P, *Q*, *R* lie on a same straight line.

1. A straight line that touches a circle at a common point is
called a _______.

2. A chord is a subsection of _______.

3. The lengths of the two tangents drawn from _______ point to a
circle are equal.

4. No tangent can be drawn from _______ of the circle.

5. _______ is a cevian that divides the angle, into two equal
halves.

Tags : Definition, Statement, Proof, Solved Example Problems | Geometry , 10th Mathematics : UNIT 4 : Geometry

Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail

10th Mathematics : UNIT 4 : Geometry : Concurrency Theorems | Definition, Statement, Proof, Solved Example Problems | Geometry

**Related Topics **

Privacy Policy, Terms and Conditions, DMCA Policy and Compliant

Copyright © 2018-2023 BrainKart.com; All Rights Reserved. Developed by Therithal info, Chennai.