$BM$ 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 $LM = LN$.
Given:
$BM$ and $CN$ are perpendiculars to a line passing, through the vertex $A$ of a triangle $ABC$.
$L$ is the mid-point of $BC$.
To do:
We have to prove that $LM = LN$.
Solution:
Join $ML$ and $NL$.
In $\triangle BMP$ and $\Delta CNP$,
$\angle \mathrm{M}=\angle \mathrm{N}$
$\angle \mathrm{BPM}=\angle \mathrm{CPN}$ (Vertically opposite angles)
Therefore, by AA similarity,
$\Delta \mathrm{BMP} \sim \Delta \mathrm{CNP}$
This implies,
$\frac{\mathrm{BM}}{\mathrm{CN}}=\frac{\mathrm{PM}}{\mathrm{PN}}$
In $\triangle \mathrm{BML}$ and $\Delta \mathrm{LNC}$,
$\frac{\mathrm{BM}}{\mathrm{CN}}=\frac{\mathrm{PM}}{\mathrm{PN}}$
$\angle \mathrm{B}=\angle \mathrm{C}$ (Alternate angles)
$\triangle \mathrm{BML} \sim \Delta \mathrm{LMC}$
This implies,
$\frac{\mathrm{ML}}{\mathrm{LN}}=\frac{\mathrm{BL}}{\mathrm{LC}}$
$\mathrm{BL}=\mathrm{LC}$
This implies,
$\frac{\mathrm{BL}}{\mathrm{LC}}=1$
$\frac{\mathrm{ML}}{\mathrm{LN}}=1$
$\mathrm{ML}=\mathrm{LN}$
Hence proved.
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