Question

if AB>AC of a triangle ABC, The bisector of angle BAC intersect at D of the side BC. Then prove that BD>CD.​

Answer 1

Answer:

i) AB > AC

II) ∠BAD = ∠CAD

To prove : BD > CD

Solution: Since AB > AC , therefore ∠ABD=∠ACD

then from angle sum property we can conclude that none of the base angles of triangles ABD and ACD are equal which means they aren’t similar (statement A).

since sum of any 2 sides is greater than the 3rd side,

therefore AD + BD > AB ————(i) and AD + CD > AC————(ii)

From statement (A) and (i) and (ii) we can conclude that AD + BD > AD + CD

which implies that BD > CD (HENCE PROVED)

Hope this helps : )

Explanation:

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