and are two isosceles triangles on the same base BC and vertices …

Question

$\triangle\text{ABC}$ and $\triangle\text{DBC}$ are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. If AD is extended to intersect BC at E, show that:

$\triangle\text{ABD}\cong \triangle\text{ACD}$
$\triangle\text{ABE}\cong\triangle\text{ACE}$
AE bisects $\angle\text{A}$ as well as $\angle\text{D}$
AE is the perpendicular bisector of BC.

✓

Answer

In $\triangle\text{ABD}$ and $\triangle\text{ACD},$

$\text{AB = AC}$ $\big($equal sides of isosceles $\triangle\text{ABC}\big)$
$\text{DB = DC}$ $\big($equal sides of isosceles $\triangle\text{DBC}\big)$
$\text{AD = AD}$ (common)
$\therefore\triangle\text{ABD}\cong\triangle\text{ACD}$ (by SSS congruence criterion)

Since $\triangle\text{ABD}\cong\triangle\text{ACD},$

$\angle\text{BAD}=\angle\text{CAD}$ (C.P.C.T.)
$\Rightarrow\angle\text{BAE}=\angle\text{CAE}...(1)$
Now, in $\triangle\text{ABE}$ and $\triangle\text{ACE}$
$\text{AB = AC}$ $\big($equal sides of isosceles $\triangle\text{ABC}\big)$
$\angle\text{BAE}=\angle\text{CAE}$ [From (1)]
$\text{AE = AE}$ (common)
$\therefore\triangle\text{ABE}\cong\triangle\text{ACE}$ (by SAS congruence criterion)

Since $\triangle\text{ABD}\cong\triangle\text{ACD},$

$\angle\text{BAD}=\angle\text{CAD}$ (C.P.C.T.)
$\Rightarrow\angle\text{BAE}=\angle\text{CAE}$
Thus, AE bisects $\angle\text{A}.$
In $\triangle\text{BDE}$ and $\triangle\text{CDE},$
$\text{BD = CD}$ $\big($equal sides of isosceles $\triangle\text{ABC}\big)$
$\text{BE = CE}$ $\big($C.P.C.T. since $\triangle\text{ABE}\cong\triangle\text{ACE}\big)$
$\text{DE = DE}$ (common)
$\therefore\triangle\text{BDE}\cong\triangle\text{CDE}$ (by SSS congruence criterion)
$\Rightarrow\angle\text{BDE}=\angle\text{CDE}$ (C.P.C.T.)
Thus, DE bisects $\angle\text{D},$ i.e., AE bisects $\angle\text{D}.$
Hence, AE bisects $\angle\text{A}$ as well as $\angle\text{D}.$

Since $\triangle\text{BDE}\cong\triangle\text{CDE},$

$\Rightarrow\text{BE = CE}$ and $\angle\text{BED}=\angle\text{CED}=90^{\circ}$ $\big($since $\angle\text{BED}$ and $\angle\text{CED}$ form a linear pair$\big)$
$\Rightarrow\text{DE}$ is the perpendicular bisector of BC.
$\Rightarrow\text{AE}$ is the perpendicular bisector of BC.