Question

$△ABC$ and $△DBC$ are two isosceles triangles on the same base $BC$ (see in fig.) If $AD$ is extended to intersect $BC$ at $P$, show that
(i) $△ABD\cong △ACD$
(ii) $△ABP\cong △ACP$
(iii) $AP$ bisects $\angle A$ as well as $\angle D$.
(iv) $AP$ is the perpendicular bisector of $BC$.

Answer 1

(i)To prove $\Delta ABD\&\Delta ACD$:
$AB=AC$(Given,isosceles)
$BD=CD$ (Given,isosceles)
$AD=AD$ (common)
By S.S.S congruency, we say that
$\Delta ABD\cong \Delta ACD$.......(1)

(ii)To prove $\Delta ABP\&\Delta ACP$:

$AB=AC$ (given)

$AP=AP$ (common)

Since, from (i) we have $\Delta ABD\cong \Delta ACD$
$\angle DAB=\angle DAC$ ( Corresponding parts of congruent triangles[C.P.C.T])
$\Rightarrow \angle BAP=\angle PAC$ ...(2)

By S.A.S property, we say that
$\Delta ABP\cong \Delta ACP$ ......(3)

(iii) Since, from(2), we have $\angle BAP=\angle PAC,$

$\therefore APbisects\angle A$

also, from (1) $\Delta ABD\cong \Delta ACD$

$\therefore \angle BDA=\angle CDA$ (C.P.C.T)

$\therefore AP\text{also}\text{bisects}\angle D$

(iv) From equation(3) $\Delta ABP\cong \Delta ACP$

$\therefore \angle APB=\angle APC$ (C.P.C.T)........(4)

and sum of angles on a straight line $={180}^{\circ }$

$\Rightarrow \angle APB+\angle APC={180}^{\circ }$
from equation(4), we get

$\Rightarrow \angle APB=\angle APC={90}^{\circ }$

$\therefore AP$ is the perpendicular bisector of$BC$