Question

$\text{AD is an altitude of an isosceles}\Delta ABC\text{in which AB = AC.}\text{Show that (i) AD bisects BC,}\text{(ii) AD bisects}\angle A.$

(2 Mark)

Answer 1

Given:-
ABC is a isosceles triangle.
AD is perpendicular on BC.
AB=BC (1 Mark)
To prove:-
In Triangle ACD AND ABD,
AD=AD (Common)
AC=AB (Given)
Angle ADC=Angle ADB (90 degree
each)
Hence,Triangle ACD=ABD (By SAS)
Therefore,AD bisects BC. (By
C.P.C.T.) (1 Mark)