Question

$ABC$ is an isosceles triangle in which altitudes $BD$ and $CE$ are drawn to equal sides $AC$ and $AB$ respectively (see figure). Show that these altitudes are equal

Answer 1

ABC is an isosceles triangle in which altitudes BD and CE are drawn to equal sides AC and AB respectively .

Given ABC is an isosceles triangle

Then AB=AC

$\therefore \angle ABC=\angle ACB$

Or $\therefore \angle EBC=\angle DCB$

In $\Delta EBC,$ $\Delta DCB$.....................(1)

$\angle BEC=\angle CDB={90}^0$

$BC=CB$

$\angle EBC=\angle DCB$

$\Delta BEC\cong \Delta CFB$.......(SAA test)

$\therefore BD=EC$

Then these altitudes are equal.