In the figure, the side BC of ΔABC is produced to D. The bisector of

1.	In the figure, the side BC of ΔABC is produced to D. The bisector of ∠A meets BC at P. Prove that: ∠ABC + ∠ACD = 2 ∠APC.
Answer» Given: The side BC of ΔABC is produced to D. The bisector of ∠A meets BC at P. To prove: ∠ABC + ∠ACD = 2∠APC Proof: ∠APC = ∠ABP + ∠BAP (the exterior angle is equal to sum of opposite interior angles) ∠APC = ∠B + \(\frac { 1 }{ 2 }\) ∠A …(i) Now in ΔABC ∠ABC + ∠BAC = ∠ACD …(ii) (reason as above) Adding ∠ABC to both side of equation (ii), we get ∠ABC + ∠ACD = ∠ABC + ∠ABC + ∠BAC = 2 ∠ABC + ∠BAC = 2 ∠B + ∠A = 2(∠B + \(\frac { 1 }{ 2 }\) ∠A) = 2 ∠APC [from equation (i)] ∠ABC + ∠ACD = 2 ∠APC.

In the figure, the side BC of ΔABC is produced to D. The bisector of ∠A meets BC at P. Prove that: ∠ABC + ∠ACD = 2 ∠APC.

Answer»

Given: The side BC of ΔABC is produced to D.

The bisector of ∠A meets BC at P.

To prove: ∠ABC + ∠ACD = 2∠APC

Proof: ∠APC = ∠ABP + ∠BAP (the exterior angle is equal to sum of opposite interior angles)
∠APC = ∠B + \(\frac { 1 }{ 2 }\) ∠A …(i)

Now in ΔABC

∠ABC + ∠BAC = ∠ACD …(ii) (reason as above)

Adding ∠ABC to both side of equation (ii), we get

∠ABC + ∠ACD = ∠ABC + ∠ABC + ∠BAC

= 2 ∠ABC + ∠BAC

= 2 ∠B + ∠A

= 2(∠B + \(\frac { 1 }{ 2 }\) ∠A)

= 2 ∠APC [from equation (i)]

∠ABC + ∠ACD = 2 ∠APC.