Question

In triangle ABC angle A =50 and BC is produced to a point D. The bisectors of angle ABC and angle ACD meet at E. Find angle E

Answer 1

Answer

25°

∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞∞

Given

In Δ ABC

∠A = 50°

BE is the bisector of ∠ABC

CE is the bisector of ∠ACD

To Find

∠E

Solution

In ΔABC,

∠A + ∠B = ∠ACD [Exterior angle Prpperty]

50° + ∠B = ∠ACD

Dividing the equation by 2

$\mathsf{\frac{50}{2}}}$ + $\frac{/_B}{2}$ = ${\mathsf{\frac{ACD}{2}}$

[Halves of equals are equal]

25° + ∠1 = ∠2

∠2 = ∠1 + 25° → 1

In Δ ECD

∠1 + ∠E = ∠2 [Exterior angle theorem]

∠2 = ∠1 + ∠E → 2

From 1 and 2

∠1 + 25° = ∠1 + ∠E

[∠1 is cancelled on both sides]

25° = ∠E

∠E = 25°