Question

Prove that the bisector of a vertical angle of an isosceles triangle is perpendicular bisector of the base.

Answer 1

Step-by-step explanation:

Please refer to the image attached for figure

Consider PQR is an  isosceles triangle such that PQ = PR and Pl is the bisector of ∠ P.

To prove :

1. ∠PLQ = ∠PLR = 90°
2. QL = LX

Proof:

In ΔPLQ and ΔPLR

PQ = PR (given)

PL = PL (common)

∠QPL = ∠RPL ( PL is the bisector of ∠P)

⇒ ΔPLQ ≅ ΔPLR                              ( SAS congruence criterion)

⇒ QL = LR                                         (by C.P.C.T)

Now,

∠PLQ + ∠PLR = 180° ( linear pair)

2∠PLQ = 180° (as ∠PLQ = ∠PLR by C.P.C.T)

⇒ ∠PLQ = 180° / 2

              = 90°  

∴ ∠PLQ = ∠PLR = 90°

Thus, ∠PLQ = ∠PLR = 90° and QL = LR.

Hence, the bisector of the vertical angle an isosceles triangle bisects the base at right angle.

Hope this helps!

If it helps then mark it as Brainliest!

Regards,

Soham Patil.

Answer 2

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.