Question

daksha was asked to paint a triangular portion of wall in three different colors . She divided the portion as shown in figure . if AB=AC and D is a point in the interior of triangle ABC such that angle DBC= angle DCB, Prove that AD bisects Angle BAC.​

Answer 1

Answer:

Please find below the solution to the asked query:

We form our diagram from given information , As given above:⬆️⬆️

As given AB  =  AC so from base angle theorem we get

∠ ABC  =  ∠ ACB                                                      --- ( 1 )

And given : ∠ DBC =  ∠ DCB                                    --- ( 2 ) from base angle theorem we get

DB  =  DC                                                                  ---- ( 3 )

We subtract equation 2 from equation 1 and get

∠ ABC -  ∠ DBC =  ∠ ACB -  ∠DCB

∠ ABD =  ∠ACD                                                  ---- ( 4 )

In ∆ ABD and ∆ ACD

AB  =  AC                                                         ( Given )

∠ ABD = ∠ ACD                                             ( From equation 4 )

DB =  DC                                                          ( From equation 3 )

So,

∆ ABD ≅ ∆ ACD                                          ( By SAS rule )

Then ,

∠ BAD  =  ∠ CAD                                        ( By CPCT )

From above equation we can say that AD is angle bisector of ∠ BAC .                             ( Hence proved )

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Answer 2

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