Question

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

Answer 1

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

To prove : ∠PLQ = ∠PLR = 90°

and QL = LX

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 cpct)

and ∠PLQ + ∠PLR = 180° ( linear pair)

2∠PLQ = 180°

∠PLQ = 180° / 2 = 90° 

∴ ∠PLQ = ∠PLR = 90°

Thus, ∠PLQ = ∠PLR = 90° and QL = LR.

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


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Answer 2

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 & ∆EDCAD= 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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Hope this will help you....