If the bisector of the vertical angle of a triangle bisects the base, show that the triangle is isosceles.

Question

Philoid · Accepted Answer

Consider ΔABC and AM is the bisector of ∠A and AM bisects the base BC so BM = CM

Extend segment AM to D such that AM = MD and join points C and D to from ΔDMC as shown

To prove ΔABC is isosceles we have to prove that AB = AC

Consider ΔAMB and ΔDMC

AM = MD … construction

∠AMB = ∠DMC … vertically opposite angles

BM = MC … AM bisects BC given

Hence by SAS test for congruency

ΔAMB ≅ ΔDMC

⇒ AB = CD … corresponding sides of congruent triangles … (i)

⇒ ∠BAM = ∠CDM … corresponding angles of congruent triangles … (a)

⇒ ∠BAM = ∠MAC … given AM is angle bisector of ∠A … (b)

Thus using (a) and (b) we can conclude that

∠CDM = ∠MAC … (c)

Now consider ΔACD

∠CAD = ∠CDA … from (c)

As the two angles are equal ΔACD is isosceles hence we can say that

⇒ AC = CD … (ii)

Now using (i) and (ii) we can conclude that

⇒ AB = AC

And hence ΔABC is isosceles triangle