Question

if the bisector of an angle of a triangle also bisects the opposite side, prove that the triangle is isosceles ​

Answer 1

HEY MATE ..

Let the triangle be ABC and let the internal bisector AD from A to the midpoint of BC.

Then we apply the sines theorem in the two triangles ABD and ADC,

which are created and we take:

BD/[sin(A/2)] = AD/(sinB) … (1)

DC/[sin(A/2)] = AD/(sinC) … (2

Now, since BD = DC, the LHS of (1) and (2) are equal,

therefore we take:

AD/(sinB) = AD/(sinC)

=> sinB = sinC … (3)

Since no angle of a triangle can be greater than or equal to 180 degrees, it follows that both angles B and C are less than 90 degrees,

therefore from (3) we conclude that they are equal. This means that the triangle ABC is isosceles.

HOPE IT HELPS YOU

HOPE IT HELPS YOU 30 THANKS + FOLLOW =INBOX