Please answer tomorrow is my exam ......  

1. In triangle ABC angle ABC and angle ACB are equal and bisectors of angle ABC and ACB meet at O find the measure of angle A  
2. In  triangle ABC the AB is produced on both sides then find the difference between the sum of exterior angles so formed by extending BC and measure of angle A
3. In triangle ABC if AB= AC and AB is produced to C such that BD =BC find ratio of angle ACD to angle ADC                                                                                                                                                                                                                                                                                                                                                    please answer ..................

3)

Given, a ΔABC in which AB is extended to D such that BD = BC

Also, AB = AC

⇒ ∠ABC = ∠ACB [Angles opposite to equal sides are equal]

Again, BD = BC

⇒ ∠BDC = ∠BCD [Angles opposite to equal sides are equal]

Now, in ΔABC by angle sum property of triangle

∠BAC + ∠ABC + ∠ACB = 180º

⇒ ∠BAC = 180º - (∠ACB + ∠ABC) = 180º - (2∠ACB)

⇒ ∠BAC = 180º - (2∠ACB)  .... (1)

Again, in ΔADC by angle sum property of triangle

∠DAC +∠ADC + ∠ACD = 180º

⇒ 180º - (2∠ACB) +∠ADC + ∠ACD = 180º  [Using (1), as ∠BAC = ∠DAC = 180º - (2∠ACB)] 

⇒ - 2∠ACB +∠ADC + ∠ACB + ∠BCD = 0 

⇒ - ∠ACB +∠ADC + ∠BCD = 0 

⇒ - ∠ACB +∠ADC + ∠ADC = 0  [As, ∠ADC =∠BDC = ∠BCD]

⇒ - ∠ACB + 2∠ADC = 0 

⇒  ∠ACB = 2∠ADC  ... (2) 

Now,

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