Question

A triangle $ABC$ has $\angle B=\angle C$. Prove that:(a)The perpendiculars from the mid-point of $BC$ to $AB$ and $AC$ are equal.(b)The perpendiculars from $B$ and $C$ to the opposite sides are equal.

Answer 1

$\left(i\right)$

$Given:-$

In $△ABC$, $\angle B=\angle C$

$DL$ is perpendicular from $D$ to $AB$

$DM$ is the perpendicular from $D$ to $AC$

$Toprove:-$

$DL=DM$

$Proof:-$

In $△DLB$ and $△DMC$

$\Rightarrow$ $\angle DLB=\angle DMC={90}^o$ [ $DL\perp AB$ and $DM\perp AC$ ]

$\Rightarrow$ $\angle B=\angle C$ [ Given ]

$\Rightarrow$ $BD=AC$ [ $D$ is the mid point of $BC$ ]

$\therefore$ $△DLB\cong △DMC$ [ By $AAS$ criteria of congruence ]

$\therefore$ $DL=DM$ [ c.p.c.t ]

$\left(ii\right)$

$Given:-$

In $△ABC$, $\angle B=\angle C.$

$BP$ is perpendicular from $D$ to $AC$

$CQ$ is the perpendicular from $C$ to $AB$

$Toprove:-$

$BP=CQ$

$Proof:-$

In $△BPC$ and $△CQB$

$\Rightarrow$ $\angle B=\angle C$ [ Given ]

$\Rightarrow$ $\angle BPC=\angle CQB={90}^o$ [ $BP\perp AC$ and $CQ\perp AB$ ]

$\Rightarrow$ $BC=BC$ [ Common side ]

$\Rightarrow$ $△BPC\cong △CQB$ [ By $AAS$ criteria ]

$\therefore$ $BP=CQ$ [ c.p.c.t ]