BM and CN are perpendicular to a line passing through the vertex A of a triangle ABC. If L is the mid-point of BC, prove that LM = LN.

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Solution

In $Δ A B C$ , BM and CN are perpendicular on a line drawn from A.

L is the mid point of BC.

ML and NL are joined.

To prove : ML = NL

Proof : In $Δ B M P a n d Δ C N P$

$∠ M = ∠ N$ (each $90^{\circ})$

$∠ B P M = ∠ C P N$ (vertically opposite angles)

$∴ Δ B M P \sim Δ C N P$ (AA criterion)

$∴ \frac{B M}{C N} = \frac{P M}{P N}$

Now in $Δ B M L a n d Δ L N C$

$\frac{B M}{C N}$ (proved)

$∠ B = ∠ C$ (alternate angles)

$(∵$ Bm and CN are perpendicular on AX)

$∴ Δ B M L \sim Δ L M C$

$∴ \frac{M L}{L N} = \frac{B L}{L C}$

But BL =LC (CL is mid-point of BC)

$∴ \frac{B L}{L C} = 1 \Rightarrow \frac{M L}{L N} = 1$

$∴ M L = L N$

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