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# In quadrilateral ACBD, AC = AD and AB bisects $$\angle{A}$$ (see figure). Show that $$\triangle{ABC}$$ $$\displaystyle \cong$$ $$\triangle{ABD}$$. What can you say about BC and BD ?

In $$\triangle{ABC}$$ and $$\triangle{ABD}$$, we have,

As, AB bisects $$\angle{A}$$,

Therefore, $$\angle{CAB}$$ = $$\angle{DAB}$$

Also, AB is a common side,

So, we can say,
$$\triangle{ABC}$$ $$\displaystyle \cong$$ $$\triangle{ABD}$$ ...(by SAS test of congruency)

Also, BC = BD ...(By CPCT)
Hence, proved.