PROVE THAT TRIANGLE AOB IA AN ISOCELES TRIANGLE!!!!!
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In ∆ADB and ∆ ACB
we have
AD = BC (given)
AC = BD (given)
AB = AB (common)
so ∆ADB congruent to ∆ACB by SSS.
So we have
angle OAB and angle OBA equal (c p c t)
Now we have
In ∆ AOB
angle OAB = angle OBA.
We know that if two angles are equal in a triangle then the opposite sides of the triangle are equal. So this means that
AO = BO.
Since two sides are equal the ∆AOB is isoceles.
Or you can write
Since two angles are equal ∆AOB is isoceles
Hope it helps dear friend ☺️
we have
AD = BC (given)
AC = BD (given)
AB = AB (common)
so ∆ADB congruent to ∆ACB by SSS.
So we have
angle OAB and angle OBA equal (c p c t)
Now we have
In ∆ AOB
angle OAB = angle OBA.
We know that if two angles are equal in a triangle then the opposite sides of the triangle are equal. So this means that
AO = BO.
Since two sides are equal the ∆AOB is isoceles.
Or you can write
Since two angles are equal ∆AOB is isoceles
Hope it helps dear friend ☺️
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