If the altitudes from two vertices of a triangle to the opposite sides are equal, then the triangle is

(b) isosceles

In $\triangle A B C$, BL $\|$ AC.

CM $\| \mathrm{AB}$ such that $\mathrm{BL}=\mathrm{CM}$.

To prove: AB =AC

In $\triangle \mathrm{ABL}$ and $\triangle \mathrm{ACM}$

$\mathrm{BL}=\mathrm{CM} \quad$ (Given)

$\angle \mathrm{BAL}=\angle \mathrm{CAM} \quad$ (Common angle)

$\angle \mathrm{ALB}=\angle \mathrm{AMC} \quad$ (Each $90^{\circ}$ )

$\triangle \mathrm{ABL} \cong \triangle \mathrm{ACM} \quad(\mathrm{AAS}$ criterion $)$

$\therefore \mathrm{AB}=\mathrm{AC} \quad(\mathrm{CPCT})$