ABCD is a cyclic quadrilateral. BA is produced to E . AD bisects EAC . Prove that DC=DB with reasons
Given, ABCD is a cyclic quadrilateral. So, ∠A + ∠C = 180° ---------(1) [because the opposite angles in a cyclic quadrilateral add up to 180°]
Also given, AD bisects angle EAC. So, ∠EAD = ∠DAC -----------(2) [because AD is the angle bisector]
Let's extend AD to meet BC at point G.
In triangle EAC, we know that: ∠E + ∠A + ∠C = 180° ---------(3) [because the sum of angles in a triangle is 180°]
Substituting the value of ∠A + ∠C from equation (1) in equation (3), we get:
∠E + 180° - ∠A - ∠C = 180°
∠E = ∠A + ∠C -----------(4)
Now, in triangle EGA, we have:
∠AGE + ∠A + ∠E = 180° ---------------[because the sum of angles in a triangle is 180°]
∠AGE = 180° - ∠A - ∠E ---------------(5)
Substituting the value of ∠E from equation (4) in equation (5), we get:
∠AGE = 180° - ∠A - (∠A + ∠C)
∠AGE = ∠C ------------(6)
Now, in triangle GCD, we have:
∠DCG + ∠D + ∠G = 180° ------------[because the sum of angles in a triangle is 180°]
Substituting the value of ∠D from equation (2) and the value of ∠AGE from equation (6), we get:
∠DCG + ∠DAC + ∠AGE = 180°
∠DCG + ∠DAC + ∠C = 180°
∠DCG + ∠C/2 + ∠C = 180° ------------- [because AD bisects ∠EAC, so ∠DAC = ∠EAD = ∠AGE]
∠DCG + 3∠C/2 = 180°
∠DCG = 180° - 3∠C/2
Now, in triangle BCG, we have:
∠BCG + ∠C + ∠G = 180° ---------------[because the sum of angles in a triangle is 180°]
Substituting the value of ∠G from equation (6), we get:
∠BCG + ∠C + ∠AGE = 180°
∠BCG + ∠C + ∠C = 180°
∠BCG + 2∠C = 180°
∠BCG = 180° - 2∠C
Since ABCD is a cyclic quadrilateral, we know that:
∠BCD + ∠BAD = 180° ---------[because the opposite angles in a cyclic quadrilateral add up to 180°]
Substituting the values of ∠BCG and ∠DAC from above, we get:
∠BCD + (∠DCG + ∠DAC) = 180°
∠BCD + (180° - 3∠C/2 + ∠C/2) = 180°
∠BCD - ∠C + 90° = 0
∠BCD = ∠C - 90°
Similarly, in triangle ABD, we have:
∠ABD + ∠BAD = 180° ---------[because the opposite angles in a cyclic quadrilateral add up to 180°]
Substituting the values of ∠AGE and ∠DAC from above, we get:
∠ABD + (∠AGE + ∠DAC) = 180°
∠ABD + (∠C + ∠C/2) = 180°
∠ABD = 180° - (3/2)∠C
Now, in triangle CDB, we have:
∠DCB + ∠BCD + ∠CBD = 180° ------------[because the sum of angles in a triangle is 180°]
Substituting the values of ∠BCD and ∠ABD from above, we get:
∠DCB + (∠C - 90°) + (∠C/2) = 180°
∠DCB + (3/2)∠C - 90° = 180°
∠DCB + (3/2)∠C = 270°
Subtracting ∠DCB from both sides, we get:
(3/2)∠C = 270° - ∠DCB
Since ∠BCD + ∠DCB = 180°, we can substitute ∠BCD = ∠C - 90° in the above equation and simplify:
(3/2)∠C = 270° - (180° - ∠C)
(3/2)∠C = 90° + ∠C
(1/2)∠C = 90°
∠C = 180°
Therefore, we have:
∠BCD = ∠C - 90° = 180° - 90° = 90°
∠BDC = 90°/2 = 45° --------------[because in an isosceles right triangle, the base angles are equal and each measure 45°]
So, we have shown that ∠BCD = 90° and ∠BDC = 45°, which implies that triangle BCD is an isosceles right triangle with DB = DC. Hence proved.