# Two isosceles tri Given that ΔABC and ΔDBC are isosceles triangles.

Here, AB=AC and DB=DC

Also, ABC=ACB

And DBC=DCB

Now, in ΔABD and ΔACD,

AB=AC (given)

BD=DC (given)

ΔABD ΔACD (by SSS rule)

Again, in ΔABO and ΔACO,

AO=AO (common)

AB=AC (given)

ΔABD ΔACD (by SAS rule)

So, BO=OC (by cpct)

And AOB=AOC(by cpct)

Also, AOB+AOC=180° (straight angle)

AOB+AOB=180°

2AOB=180°

AOB=90°

So, AOC=AOB=90°

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