In $△EFG$, the line segment $EH$ is the perpendicular bisector of the line segment $FG$. Prove that $△EFH\cong △EGH$ using $\mathrm{SSS}$ congruence rule.

In $∆ABC, \angle A=58°$. If $I$ is the incentre of the triangle, then the measure of $\angle BIC$ is _____.

In the given figure, $△\mathrm{ABC}$ is an isosceles triangle, in which $\mathrm{AB}=\mathrm{AC}, \mathrm{AD}\perp \mathrm{BC}, \mathrm{BC}=6\mathrm{cm}$ and $\mathrm{AD}=4 \mathrm{cm}$. The length of $\mathrm{AB}$ is:

In triangle $ABC$, $AD$ is the angle bisector of $\angle BAC$. If $AB=c, AC=b$ and $BC=a$, then the measure of the segment $BD$ is equal to:

$ABC$ is a triangle. The bisectors of the internal angle $\angle B$ and external angle $\angle C$ intersect at $D.$ If $\angle BDC = {50}^{\mathrm{o}}$ then $\angle A$ is -

In the given figure, $ABC$ is a triangle. The bisectors of internal $\angle B$ and external $\angle C$ ntersect at $D$. If $\angle BDC={48}^{\mathrm{o}}$, then what is the value (in degrees) of $\angle A$?

$ABC$ is a triangle. $AD$ is perpendicular to $BC$ and $AE$ is the bisector of the angle $BAC$. The measure of the angle $DAE$ is equal to: