English

Given below is the triangle and length of line segments. Identify in the given figure, ray PM is the bisector of ∠QPR. - Geometry Mathematics 2

Advertisements
Advertisements

Question

Given below is the triangle and length of line segments. Identify in the given figure, ray PM is the bisector of ∠QPR.

Sum

Solution

In ΔPQR,

`"PR"/"PQ" = 7/10`    .....(i)

`"RM"/"QM" = 6/8`   ......(ii)

From (i) and (ii),

`"PR"/"PQ" ≠ "RM"/"QM"`

∴ Ray PM is not the bisector of ∠QPR.

shaalaa.com
Property of an Angle Bisector of a Triangle
  Is there an error in this question or solution?
Chapter 1: Similarity - Practice Set 1.2 [Page 13]

APPEARS IN

RELATED QUESTIONS

Given below is the triangle and length of line segments. Identify in the given figure, ray PM is the bisector of ∠QPR.


Measures of some angles in the figure are given. Prove that `"AP"/"PB" = "AQ"/"QC"`.


Find QP using given information in the figure.


In the given figure, if AB || CD || FE then find x and AE. 


In ∆LMN, ray MT bisects ∠LMN If LM = 6, MN = 10, TN = 8, then Find LT. 


In ∆ABC, seg BD bisects ∠ABC. If AB = x, BC = x + 5, AD = x – 2, DC = x + 2, then find the value of x.


In ∆PQR seg PM is a median. Angle bisectors of ∠PMQ and ∠PMR intersect side PQ and side PR in points X and Y respectively. Prove that XY || QR. 


Complete the proof by filling in the boxes.

In △PMQ, ray MX is bisector of ∠PMQ.

∴ `square/square = square/square` .......... (I) theorem of angle bisector.

In △PMR, ray MY is bisector of ∠PMQ.

∴ `square/square = square/square` .......... (II) theorem of angle bisector.

But `(MP)/(MQ) = (MP)/(MR)` .......... M is the midpoint QR, hence MQ = MR.

∴ `(PX)/(XQ) = (PY)/(YR)`

∴ XY || QR .......... converse of basic proportionality theorem.


In the given fig, bisectors of ∠B and ∠C of ∆ABC intersect each other in point X. Line AX intersects side BC in point Y. AB = 5, AC = 4, BC = 6 then find `"AX"/"XY"`.


Seg NQ is the bisector of ∠ N
of Δ MNP. If MN= 5, PN =7,
MQ = 2.5 then find QP.


From the top of a light house, an abserver looking at a boat makes an angle of depression of 600. If the height of the lighthouse is 90 m then find how far is the boat from the lighthouse. (3 = 1.73)


In ΔABC, ray BD bisects ∠ABC.

If A – D – C, A – E – B and seg ED || side BC, then prove that:

`("AB")/("BC") = ("AE")/("EB")`

Proof : 

In ΔABC, ray BD bisects ∠ABC.

∴ `("AB")/("BC") = (......)/(......)`   ......(i) (By angle bisector theorem)

In ΔABC, seg DE || side BC

∴ `("AE")/("EB") = ("AD")/("DC")`   ....(ii) `square`

∴ `("AB")/square = square/("EB")`   [from (i) and (ii)]


In the figure, ray YM is the bisector of ∠XYZ, where seg XY ≅ seg YZ, find the relation between XM and MZ. 


Draw seg AB = 6.8 cm and draw perpendicular bisector of it. 


Draw the circumcircle of ΔPMT in which PM = 5.6 cm, ∠P = 60°, ∠M = 70°.


From the information given in the figure, determine whether MP is the bisector of ∠KMN.



In ∆PQR seg PM is a median. Angle bisectors of ∠PMQ and ∠PMR intersect side PQ and side PR in points X and Y respectively. Prove that XY || QR. 

Complete the proof by filling in the boxes.

solution:

In ∆PMQ,

Ray MX is the bisector of ∠PMQ.

∴ `("MP")/("MQ") = square/square` .............(I) [Theorem of angle bisector]

Similarly, in ∆PMR, Ray MY is the bisector of ∠PMR.

∴ `("MP")/("MR") = square/square` .............(II) [Theorem of angle bisector]

But `("MP")/("MQ") = ("MP")/("MR")`  .............(III) [As M is the midpoint of QR.] 

Hence MQ = MR

∴ `("PX")/square = square/("YR")`  .............[From (I), (II) and (III)]

∴ XY || QR   .............[Converse of basic proportionality theorem]


In ΔABC, ray BD bisects ∠ABC, A – D – C, seg DE || side BC, A – E – B, then for showing `("AB")/("BC") = ("AE")/("EB")`, complete the following activity:

Proof :

In ΔABC, ray BD bisects ∠B.

∴ `square/("BC") = ("AD")/("DC")`   ...(I) (`square`)

ΔABC, DE || BC

∴ `(square)/("EB") = ("AD")/("DC")`   ...(II) (`square`)

∴ `("AB")/square = square/("EB")`   ...[from (I) and (II)]


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×