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

Advertisements
Solution
\[\text{In} \bigtriangleup \text{LNM}, \]
\[\frac{\text{LT}}{\text{NT}} = \frac{\text{LM}}{\text{NM}} \left( \text{ By angle bisector theorem } \right)\]
\[ \Rightarrow \frac{\text{LT}}{8} = \frac{6}{10}\]
\[\Rightarrow \text{LT} = \frac{8 \times 6}{10}\]
\[ = 4 . 8\]
Hence, the measure of LT is 4.8.
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.

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

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

In ∆MNP, NQ is a bisector of ∠N. If MN = 5, PN = 7 MQ = 2.5 then find QP.

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

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

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


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

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 ΔABC, ∠ACB = 90°. seg CD ⊥ side AB and seg CE is angle bisector of ∠ACB.
Prove that: `(AD)/(BD) = (AE^2)/(BE^2)`.
In the figure, ray YM is the bisector of ∠XYZ, where seg XY ≅ seg YZ, find the relation between XM and MZ.

In the following figure, ray PT is the bisector of ∠QPR Find the value of x and perimeter of ∠QPR.

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 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)]
