\(\triangle{ABC}\) and \(\triangle{DBC}\) are two isosceles triangles 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
(i) \(\triangle{ABD}\) \(\displaystyle \cong \) \(\triangle{ACD}\)
(ii)\(\triangle{ABP}\) \(\displaystyle \cong \) \(\triangle{ACP}\)
(iii) AP bisects \(\angle{A}\) as well as \(\angle{D}\)
(iv) AP is the perpendicular bisector of BC.

Given: \(\triangle{ABC}\) and \(\triangle{DBC}\) are two isosceles triangles having same base BC, such that AB = AC and BD = CD.

Proof:
In \(\triangle{ABD}\) and \(\triangle{ACD}\), we have,

AB = AC ...(Given)
BD = CD ...(Given)
and AD is Common side.

Therefore, \(\triangle{ABD}\) \(\displaystyle \cong \) \(\triangle{ACD}\) ...(By SSS Congruency test)
Hence, part (i) is proved.


In \(\triangle{ABP}\) and \(\triangle{ACP}\), we have,

AB = AC ...(Given)
Also, \(\angle{a}\) = \(\angle{b}\) ...(\(\because \) \(\triangle{ABD}\) \(\displaystyle \cong \) \(\triangle{ACD}\))
and AP is the common side.

Therefore, \(\triangle{ABP}\) \(\displaystyle \cong \) \(\triangle{ACP}\) ...(By SAS congruency test)
Hence, part (ii) is proved.


\(\angle{a}\) = \(\angle{b}\) ...(\(\because \) \(\triangle{ABD}\) \(\displaystyle \cong \) \(\triangle{ACD}\))
Hence, proved that AP bisects \(\angle{A}\)

Now, \(\angle{ADB}\) = \(\angle{ACD}\) ...(\(\because \)\(\triangle{ABD}\) \(\displaystyle \cong \) \(\triangle{ACD}\))

Now, by subtracting from \(180^\circ\), we get,

\(180^\circ\) - \(\angle{ADB}\) = \(180^\circ\) - \(\angle{ACD}\)
\(\Rightarrow \) \(\angle{BDP}\) = \(\angle{ACD}\) ...(By linear pair axiom)

Hence, proved that AP bisects \(\angle{D}\)

Hence, part (iii) is proved.

Now, BP = CP (Since, \(\triangle{ABP}\) \(\displaystyle \cong \) \(\triangle{ACP}\))
and \(\angle{c}\) = \(\angle{d}\) ...(i)
But \(\angle{c}\) + \(\angle{d}\) = \(180^\circ\) ...(Linear pair axiom)
\(\Rightarrow \)\(\angle{c}\) + \(\angle{c}\) = \(180^\circ\) ...(from (i))
\(\Rightarrow \) 2 \(\angle{c}\) = \(180^\circ\)
\(\Rightarrow \) \(\angle{c}\) = \(90^\circ\)
\(\Rightarrow \)\(\angle{c}\) = \(\angle{d}\) = \(90^\circ\)

Therefore, AP is the perpendicular bisector of BC.
Hence, proved.