show that the bisectors of the base angles of a triangle can never enclose a right angle.
Answers
Step-by-step explanation:
In ΔABC, BP and CP are bisectors of angles B and C respectively. Hence ∠A + ∠B + ∠C = 180° ∠A + 2∠1 + 2∠2 = 180° 2(∠1 + ∠2) = 180° – ∠A (∠1 + ∠2) = 90° – (∠A/2) --- (1) In ΔPBC, ∠P + ∠1 + ∠2 = 180° ∠P + [90° – (∠A/2)] = 180° [From (1)] ∠P = 180° – [90° – (∠A/2)] = [90° + (∠A/2)] Hence angle P is always greater than 90°. Thus PBC can never be a right angled triangle.
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Step-by-step explanation:
Let BP and CP are bisector of angles B and C respectively
we need to prove that bisectors of the base angles of a triangle can never enclose a right angle.
In ∆ABC, ∠A+∠B+∠C = 180°
Now, ∠A+ 2∠1 + 2∠C = 180°
2(∠1+∠2) = 180°-∠A
⇒∠1+∠2= 90° −
In ∆PBC,∠P+∠1+∠2=180°
∠P+90°- = 180°
∠P = 90°+
Hence angle P is always greater then 90°
Thus, PBC can never be a right angled triangle