Question

If the equal sides of an isosceles triangle are produced, prove that the exterior angles so formed are obtuse and equal.

Answer 1

Const: AB is produced to D and AC is produced to E so that exterior angles ∠DBC and ∠ECB are formed.

In ΔABC,
AB = AC ........[ Given ] 
∴ ∠C = ∠B .....(i) [angels opp. to equal sides are equal]

Since angle B and angle C are acute they cannot be right angles or obtuse angles.

∠ABC + ∠DBC =180° .......[ABD is a st. line] 
∠DBC = 180° − ∠ABC
∠DBC = 180° − ∠B ......(ii)

Similarly,
∠ACB + ∠ECB = 180° .......[ABD is a st. line] 
∠ECB = 180° − ∠ACB
∠ECB = 180° − ∠C ........(iii)
∠ECB = 180° − ∠B .......(iv) [from(i) and (iii)]
∠DBC = ∠ECB ........[from (ii) and (iv)]

Now,
∠DBC = 180° − ∠B 
But ∠B = Acute angel
∴  ∠DBC = 180° − Acute angle = obtuse angle

Similarly,
∠ECB = 180° − ∠C.
But ∠C = Acute angel
∴  ∠ECB = 180° − Acute angle = obtuse angle

Therefore, exterior angles formed are obtuse and equal.