Question

Question 1 (iv)
$\Delta$ABC and $\Delta$DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see the given figure). If AD is extended to intersect BC at P, show that
AP is the perpendicular bisector of BC.

Answer 1

In $\Delta ABD$ and $\Delta ACD$,
AB = AC (given)
BD = CD (given)
AD = AD (common)
$\therefore \Delta ABD\cong \Delta ACD$ (by SSS congruence rule)
$\Rightarrow \angle BAD=\angle CAD$ (by CPCT)
$\Rightarrow \angle BAP=\angle CAP\dots \left(i\right)$ (by CPCT)

In $\Delta ABP$ and $\Delta ACP$,
AB = AC (given).
$\angle BAP$ = $\angle CAP$ [From (i)]
AP = AP (common)
$\therefore \Delta ABP\cong \Delta ACP$ (by SAS congruence rule)

$\Rightarrow BP=CP\dots \left(ii\right)$(by C.P.C.T)
$\Rightarrow \angle APB=\angle APC\dots \left(iii\right)$(by C.P.C.T)

BC is a straight line.
Now, $\angle BPD+\angle CPD={180}^{\circ }$ (linear pair angles)
$\angle APB+\angle APC={180}^{\circ }$
$2\angle APB={180}^{\circ }$ [from equation (iii)]
$\angle APB={90}^{\circ }\dots \left(iv\right)$
From equations (ii) and (iv), we can say that AP is the perpendicular bisector of BC.