Question

prove that the bisector of the Vertical angle of an isosceles triangle is perpendicular to the base

Answer 1

Let in triangle ABC,AD is bisector of vertical angle.
In ∆ABD and ∆ACD
AB=BC
angle BAD=angle CAD
AD is common line.
By SAS congruency ∆ABD congruent to∆CBD
BD=CD
angle ABD=angle ACD
angle ADB=angle ADC
let angle ADB=x.
so angle ADC=x.
x+x=180
2x=180
x=90
Hence it's proved.

Answer 2

To Prove that the bisector of the vertical angle of an isosceles Δ is perpendicular of the base.

Explanation:

∠PCQ = ∠PLR = 90°

QL = LR

In ∆PLQ and ∆PLR

PQ = PR (given)
PL  = PL(common)

∠QPL = ∠RPL (PL is bisector of LP)

ΔPLQ = ΔPLR (SAS congruences)

QL = LR

and ∠PLQ + ∠PLR = 180°

= ∠PLQ = 180°

∠PLQ = 180°/2 = 90°

∴Hence Proved.