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In ∆Lmn, Ray Mt Bisects ∠Lmn If Lm = 6, Mn = 10, Tn = 8, Then Find Lt. - Geometry Mathematics 2

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प्रश्न

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

बेरीज

उत्तर

\[\text{In} \bigtriangleup \text{LNM}, \]
\[\frac{\text{LT}}{\text{NT}} = \frac{\text{LM}}{\text{NM}} \left( \text{ By angle bisector theorem } \right)\]
\[ \Rightarrow \frac{\text{LT}}{8} = \frac{6}{10}\] 

\[\Rightarrow \text{LT} = \frac{8 \times 6}{10}\]
\[ = 4 . 8\]

Hence, the measure of LT is 4.8.

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Property of an Angle Bisector of a Triangle
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पाठ 1: Similarity - Practice Set 1.2 [पृष्ठ १५]

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बालभारती Geometry (Mathematics 2) [English] 10 Standard SSC Maharashtra State Board
पाठ 1 Similarity
Practice Set 1.2 | Q 8 | पृष्ठ १५

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


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"`.


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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]


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