Question

in triangleABC,AB=AC and B-C-D prove that ADis smaller than AB​

Answer 1

Answer:

Step-by-step explanation:

In any triangle, the side opposite the greater angle is greater than the side opposite the smaller one. This is a consequence of the Law of Sines, and the fact that the sum of the angles of a triangle is  π .

Now, in  ΔABC , as  AB>AC ,  ∠ACD>∠ABD.  

Now, in  ΔACD , the exterior angle  ∠ADB=∠ACD+∠CAD  (sum of interior opposite angles) >  ∠ABD+∠CAD>∠ABD .

In  ΔADB,∠ADB>∠ABD . Therefore  AB>AD .

Answer 2

Answer:

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