Question

In $\Delta ABC$, the bisector AX of $\angle A$ intersects BC at X. $XL\perp AB$ and $XM\perp AC$ are drawn. Is XL = XM?
Why or why not?

Answer 1

Given : $AX$ is the bisector of $\angle A$ meet $BC$ on $X$

And $XL\perp AB$ and $XM\perp AC$

Then, in $\Delta ALX$ and $\Delta AMX$

$\angle LAX=\angle XAM$ ...... ($AX$ is the bisector of angle A)

$\angle ALX=\angle AMX={90}^o$ ...... ($XL\perp AB$ and $XM\perp AC$)

$AX=AX$ ....... [Common side]

Then, $\Delta ALX\simeq \Delta AMX$ ...... [By ASA rule of congruence]

$\therefore LX=MX$