Question

$△ABC$ and $△DBC$ are two isosceles triangle on the same base $BC$ and vertices $A$ and $D$ are on the same side of $BC$ (see figure). If $AD$ is extended to intersect $BC$ at $P$ show that $AP$ is the perpendicular bisector of $BC$

Answer 1

In $△ABC,AB=AC--\left(1\right)$ [sides of an isosceles triangle are equal]

In $△DBC,DB=DC--\left(2\right)$ [sides of an isosceles triangle are equal]

In $△ABD$ and $△ACD$

$AB=AC$ [From $\left(1\right)$]

$DB=DC$ [From $\left(2\right)$]

$AD=AD$ [Common side]

$△ABD\cong △ACD$

[by $SSS$ Test of congruency ]

$\therefore \angle BAD=\angle CAD\left(1\right)-$ [corresponding air of congurent triangle]

In $△ABP$ and $△ACP$

$AB=AC$- [From $\left(1\right)$]

$AP=AP$- [Common side]

$\angle BAP=\angle CAP$- [From $\left(3\right)$]

$△ABP\cong △ACP$- [by $SAS$ test of congruency]

$△BP=CP---\left(4\right)$ [corresponding pairs of congruency triangles]

$\angle APB=\angle APL--\left(5\right)$ [corresponding pairs of congruency triangles]

$iii]$ In $△BDP$ and $△CDP$

$DB=DC$- [From $\left(1\right)$]

$BP=CP$- [From $\left(2\right)$]

$DP=DP$- [Common side]

$\therefore △BDP\cong △CDP$- [by $SSS$ test of congruency]

$\therefore \angle BDP=\angle CDP$- [corresponding pair of congruent triangles]

Since $\angle BDP=\angle CDP$ we can say that

$AP$ bisects $\angle D$

Since, $\angle BAD=\angle CAD$- [From $\left(3\right)$]

we can say that $AP$ bisects $\angle A$

$iv]\angle APB+\angle APC={180}^o$- [Angle in a straight line is ${180}^o$]

$\therefore \angle APB+\angle APB={180}^o$- [From $\left(5\right)$]

$2\angle APB={180}^o$

$\angle APB={90}^o$

$\angle APB=\angle APC={90}^o$

From, the above result, we can say that $AP$ is the perpendicular bisector if $BC$