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