Question

if a bisector of a Vertical angle of a triangle bisect the base of a triangle then it is an isosceles triangle prove

Answer 1

Given:

In ∆ABC ,

AD bisects ∠BAC, & BD= CD

To Prove:

AB=AC

Construction:

Produce AD to E such that AD=DE & then join E to C.

Proof:

In ∆ADB & ∆EDC

AD= ED ( by construction)

∠ADB= ∠EDC. (vertically opposite angles (

BD= CD (given)

∆ADB congruent ∆EDC (by SAS)

Hence, ∠BAD=∠CED......(1) (CPCT)

∠BAD=∠CAD......(2). (given)

From eq.1 &2

∠CED =∠CAD......(3)

AB=CE (CPCT).......(4)

From eq 3 as proved that

∠CED=∠CAD

So we can say CA=CE......(5)

[SIDES OPPOSITE TO EQUAL ANGLES ARE EQUAL]

Hence, from eq 4 & 5

AB = AC

HENCE THE ∆ IS ISOSCELES..

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Answer 2

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

Assumption

∆PQR is Isosceles triangle

PQ = PR

Also,

Line PS bisect ∠QPR such that

∠QPS = ∠RPS

To Prove,

PS ⊥ QR also QS = SR

$\Large{\boxed{\sf\:{Proof :- }}}$

In ∆PQS and ∆ PRS

PQ = PR (Given)

∠QPS = ∠RPS (Given)

PS = PS (Common)

∆PQS ≅ ∆PRS (By SAS rule)

$\Large{\boxed{\sf\:{Therefore, }}}$

QS = SR (CPCT)

Also,

∠PSQ = ∠PSR (CPCT)

But,

∠PSR + ∠PSR = 180

∠PSR + ∠PSR = 90°

$\Large{\boxed{\sf\:{Therefore, }}}$

PS ⊥ QR

Hence,

PS ⊥ QR also QS = SR