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# AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that (i) AD bisects BC (ii) AD bisects $$\angle{A}$$

In $$\triangle{ABD}$$ and $$\triangle{ACD}$$, we have,
AB = AC ...(Given)

$$\angle{ABD}$$ = $$\angle{ADC}$$ = $$90^\circ$$ ...(Since AD is an altitude)
Therefore, $$\triangle{ABD}$$ $$\displaystyle \cong$$ $$\triangle{ACD}$$ ...(By RHS congruency axiom)
Also, $$\angle{BAD}$$ = $$\angle{CAD}$$ ...(By CPCT)
Thus, this shows that AD bisects $$\angle{A}$$.