Premium Online Home Tutors
3 Tutor System
Starting just at 265/hour
Answer :
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.