Question

plzz solve this question..??​

Answer 1

Answer:

Tum konsi class mein parti ho ??

Answer 2

STEP BY STEP ANSWER

(1)

from triangle ∆ABD and ∆ACD,we get

AD=AD(common sides)

AB=AC(as ∆ABC is a isosceles triangle,)

BD=DC(as ∆BDC is a isosceles triangle,)

so,

∆ABD=~∆ACD(SSS congruency)(proved)...(I)

therefore,<BAD=<CAD(cpct)

or,<BAE=<CAE

i.e., AD bisects <A .

∆ABC is a isosceles triangle,and AD bisects <A .

so AD or AE is the median of ∆ABC,

so,BE=CE

(2)

from triangle ∆ABE and ∆ACE,we get

AE=AE(common sides)

<BAE=<CAE(proved earlier)

AB=AC(as ∆ABC is a isosceles triangle,)

so,

∆ABE=~∆ACE(SAS congruency)(proved)

(3)

from triangle ∆BDEand ∆CDE,we get

DE=DE(common sides)

BE=CE(proved earlier,)

BD=DC(as ∆BDC is a isosceles triangle,)

so,

∆BDE=~∆CDE(SAS congruency)(proved)

so,

<BDE=<CDE(cpct)

i.e., DE bisects <D or AE bisects <D (proved)

and from previous

or,<BAE=<CAE

i.e., AD bisects <A or AE bisects <A (proved)

(4)

now ,∆BDE=~∆CDE

so <DEB=<DEC(cpct)

<DEB+<DEC=2 right angle

2<DEB=2 right angle

<DEB=1 Right angle

i.e.,DE perpendicular to BC

or AE perpendicular to BC(proved)