In a $\triangle ABC, BM$ and $CN$ are perpendiculars from $B$ and $C$ respectively on any line passing through $A$. If $L$ is the mid-point of $BC$, prove that $ML=NL$.
Given:
In a $\triangle ABC, BM$ and $CN$ are perpendiculars from $B$ and $C$ respectively on any line passing through $A$.
$L$ is the mid-point of $BC$.
To do:
We have to prove that $ML=NL$.
Solution:
Join $ML$ and $NL$.
In $\triangle BMP$ and $\triangle CNP$,
$\angle M=\angle N=90^o$
$\angle BPM=\angle CPN$ (Vertically opposite angles)
Therefore, by AA axiom,
$\Delta \mathrm{BML} \sim \triangle \mathrm{LMC}$
This implies,
$\frac{BM}{CN}=\frac{PM}{PN}$
In $\triangle BML$ and $\triangle CNL$,
$\frac{BM}{CN}=\frac{PM}{PN}$
$\angle B=\angle C$ (Alternate angles)
Therefore,
$\triangle BML \sim \triangle LNC$.
This implies,
$\frac{ML}{LN}=\frac{BL}{LC}$
$\mathrm{BL}=\mathrm{LC}$
This implies,
$\mathrm{CL}$ is the mid point of $\mathrm{BL}$
$\frac{\mathrm{BL}}{\mathrm{LC}}=1$
$\Rightarrow \frac{\mathrm{ML}}{\mathrm{LN}}=1$
Therefore,
$\mathrm{ML}=\mathrm{LN}$.
Hence proved.
Related Articles
- $BM$ and $CN$ are perpendiculars to a line passing, through the vertex $A$ of a triangle $ABC$. If $L$ is the mid-point of $BC$, prove that $LM = LN$.
- $ABC$ is a triangle and $D$ is the mid-point of $BC$. The perpendiculars from $D$ to $AB$ and $AC$ are equal. Prove that the triangle is isosceles.
- In a $\triangle ABC$, if $L$ and $M$ are points on $AB$ and $AC$ respectively such that $LM \| BC$.Prove that $ar(\triangle LCM) = ar(\triangle LBM)$.
- In a $\triangle ABC$, if $L$ and $M$ are points on $AB$ and $AC$ respectively such that $LM \| BC$.Prove that $ar(\triangle LBC) = ar(\triangle MBC)$.
- In a $\triangle ABC$, if $L$ and $M$ are points on $AB$ and $AC$ respectively such that $LM \| BC$.Prove that $ar(\triangle ABM) = ar(\triangle ACL)$.
- In a $\triangle ABC$, if $L$ and $M$ are points on $AB$ and $AC$ respectively such that $LM \| BC$.Prove that $ar(\triangle LOB) = ar(\triangle MOC).
- $D$ is the mid-point of side $BC$ of $\triangle ABC$ and $E$ is the mid-point of $BD$. If $O$ is the mid-point of $AE$, prove that $ar(\triangle BOE) = \frac{1}{8} ar(\triangle ABC)$.
- In $\triangle ABC$, AL and CM are the perpendiculars from the vertices A and C to BC and AB respectively. If AL and CM intersect at O, prove that: $\triangle OMA \sim OLC$.
- In a $\triangle ABC, P$ and $Q$ are respectively, the mid-points of $AB$ and $BC$ and $R$ is the mid-point of $AP$. Prove that \( \operatorname{ar}(\Delta \mathrm{PBQ})=a r(\triangle \mathrm{ARC}) \).
- $ABC$ is a triangle in which $BE$ and $CF$ are respectively, the perpendiculars to the sides $AC$ and $AB$. If $BE = CF$, prove that $\triangle ABC$ is isosceles.
- In a $\triangle ABC, P$ and $Q$ are respectively, the mid-points of $AB$ and $BC$ and $R$ is the mid-point of $AP$. Prove that \( \operatorname{ar}(\Delta \mathrm{RQC})=\frac{3}{8} \operatorname{ar}(\triangle \mathrm{ABC}) \).
- If $D$ and $E$ are points on sides $AB$ and $AC$ respectively of a $\triangle ABC$ such that $DE \parallel BC$ and $BD = CE$. Prove that $\triangle ABC$ is isosceles.
- $ABC$ is a triangle and through $A, B, C$ lines are drawn parallel to $BC, CA$ and $AB$ respectively intersecting at $P, Q$ and $R$. Prove that the perimeter of $\triangle PQR$ is double the perimeter of $\triangle ABC$.
- In a $\triangle ABC, P$ and $Q$ are respectively, the mid-points of $AB$ and $BC$ and $R$ is the mid-point of $AP$. Prove that \( \operatorname{ar}(\mathrm{PRQ})=\frac{1}{2} \operatorname{ar}(\triangle \mathrm{ARC}) \).
- In $\triangle ABC$, AL and CM are the perpendiculars from the vertices A and C to BC and AB respectively. If AL and CM intersect at O, prove that: $\frac{OA}{OC}=\frac{OM}{OL}$.
Kickstart Your Career
Get certified by completing the course
Get Started
Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies