Question

In the figure AB=AD;BC=CD. Prove that AC is the common angle bisector of angle BDA and BCD.

Answer 1

Answer:

We have ∠1=∠2 and ∠3=∠4

⇒∠1+∠3=∠2+∠4

⇒∠ACD=∠BDC.

Thus in triangles ACD and BDC, we have,

∠ADC=∠BCD   (given);

  CD  =   CD     (common);

∠ACD=∠BDC   (proved).

By ASA condition △ACD≅△BDC. Therefore

AD=BC and ∠A=∠B.

Step-by-step explanation:

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