Given that ABC and DBC are two isosceles triangles on the same base BC, this means that AB = AC and DB = DC.

In triangle ABC, since AB = AC (given), we have ∠ACB = ∠ABC (angles opposite to equal sides are equal).

Similarly, in triangle DBC, since DB = DC (given), we have ∠DBC = ∠DCB (angles opposite to equal sides are equal).

Now, ∠ABD = ∠ABC + ∠CBD (exterior angle of a triangle is equal to the sum of its interior opposite angles).

And ∠ACD = ∠ACB + ∠BCD (exterior angle of a triangle is equal to the sum of its interior opposite angles).

But we know that ∠ABC = ∠ACB and ∠CBD = ∠BCD.

Therefore, ∠ABD = ∠ACD.

Question

Given that ABC and DBC are two isosceles triangles on the same base BC, this means that AB = AC and DB = DC.

In triangle ABC, since AB = AC (given), we have ∠ACB = ∠ABC (angles opposite to equal sides are equal).

Similarly, in triangle DBC, since DB = DC (given), we have ∠DBC = ∠DCB (angles opposite to equal sides are equal).

Now, ∠ABD = ∠ABC + ∠CBD (exterior angle of a triangle is equal to the sum of its interior opposite angles).

And ∠ACD = ∠ACB + ∠BCD (exterior angle of a triangle is equal to the sum of its interior opposite angles).

But we know that ∠ABC = ∠ACB and ∠CBD = ∠BCD.

Therefore, ∠ABD = ∠ACD.

Knowee AI · Accepted Answer

Given that ABC and DBC are two isosceles triangles on the same base BC, this means that AB = AC and DB = DC.

In triangle ABC, since AB = AC (given), we have ∠ACB = ∠ABC (angles opposite to equal sides are equal).

Similarly, in triangle DBC, since DB = DC (given), we have ∠DBC = ∠DCB (angles opposite to equal sides are equal).

Now, ∠ABD = ∠ABC + ∠CBD (exterior angle of a triangle is equal to the sum of its interior opposite angles).

And ∠ACD = ∠ACB + ∠BCD (exterior angle of a triangle is equal to the sum of its interior opposite angles).

But we know that ∠ABC = ∠ACB and ∠CBD = ∠BCD.

Therefore, ∠ABD = ∠ACD.

In fig., ABC and DBC are two isosceles triangles on the same base BC.Show that ∠ABD=∠ACD.