Question

Sides BC, AB and CA of ΔABC are extended to D, E and F respectively. CG, BG, CH, AH, BI and AI are bisectors of ∠ACD, ∠ABC, ∠ACB, ∠BAF, ∠CBE and ∠BAC respectively, then sum of angles ∠BGC, ∠AHC and ∠AIB is

• A. 90°
• B. 180°
• C. 360°
• D. 120°

Answer 1

The correct option is A 90°


Given that CG, BG, CH, AH, BI and AI are bisectors of ∠ACD,
∠ABC, ∠ACB, ∠BAF, ∠CBE, ∠BAC respectively.
In ΔABC, by Angle sum property,
2∠1 + 2∠2 + 2∠3 = 180° (Using figure)
⇒ ∠1 + ∠2 + ∠3 = 90° …..(i)
In ΔABI,
(2∠2 + ∠4) + ∠3 + ∠AIB = 180° …..(ii)
In ΔBCG,
(2∠1 + ∠5) + ∠2 + ∠BGC = 180° …..(iii)
In ΔCHA,
(2∠3 + ∠6) + ∠1 + ∠AHC = 180° …..(iv)
Adding (ii), (iii) and (iv), we get
2(∠1 + ∠2 + ∠3) + (∠4 + ∠5 + ∠6) + (∠1 + ∠2 + ∠3) + ∠AIB + ∠BGC + ∠AHC = 3 × 180°
Now, let ∠AIB + ∠BGC + ∠AHC = x.
⇒ 3 × 90° + x + (∠4 + ∠5 + ∠6) = 540° (From (i))
⇒ x + (∠4 + ∠5 + ∠6) = 270° …..(v)
In ΔABC, by exterior angle property,
∠BAC + ∠ACB = ∠CBE
⇒ 2∠3 + 2∠1 = 2∠4
⇒ ∠4 = ∠1 + ∠3
Similarly, ∠5 = ∠2 + ∠3 and, ∠6 = ∠1 + ∠2
∴ ∠4 + ∠5 + ∠6 = 2(∠1 + ∠2 + ∠3)
= 2 × 90° [From (i)]
= 180°
Substituting in (v), we get
x + 180° = 270°
⇒ x = 90°
Thus, ∠BGC + ∠AHC + ∠AIB is 90°.
Hence, the correct answer is option (a).