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# $$\triangle{ABC}$$ 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)

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.