Question

$△\mathrm{ABC}$is a triangle in which altitudes $BE$ and$CF$ to sides $AC$and $AB$ are equal (see Fig.) . Show that $\mathrm{\Delta ABE}\cong \mathrm{\Delta ACF}$

Answer 1

Showing that $\mathrm{\Delta ABE}\cong \mathrm{\Delta ACF}$:

Given:

Altitudes $\mathrm{BE}$ and $\mathrm{CF}$ to sides $AC$ and $AB$ are equal

To Prove:

$\mathrm{\Delta ABE}\cong \mathrm{\Delta ACF}$

Proof:

In $∆\mathrm{ABF}$ and $∆\mathrm{ACF}$,

$\angle \mathrm{E}=\angle \mathrm{F}$ [Each $90°$ angle]

$\angle \mathrm{A}=\angle \mathrm{A}$ [common angle]

$\mathrm{AB}=\mathrm{AC}$ [given]

$\therefore ∆\mathrm{AEB}\cong ∆\mathrm{AFC}$ [ by$ASA$ criteria]

Hence Proved.