Question

In triangle ABC, the side AB is greater than the side AC . AD bisecting the angle BAC meets BC at D. Prove that BD is greater than CD.​

Answer 1

Answer:

Given:

i) AB > AC

II) /angleBAD = /angleCAD

To prove : BD > CD

Solution: Since AB > AC , therefore angle/ABD=/angleACD

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 you : )

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