Question

ABC is a right triangle with AB = AC. If bisector of ∠A meets BC at D and AD = 2√2 cm, then then perimeter

of ∆ABC is​

Answer 1

Step-by-step explanation:

search-icon-header

Search for questions & chapters

search-icon-image

Question

Bookmark

ABC is a right triangle with AB=AC. Bisector of ∠A meets BC at D. Prove that BC=2AD.

Medium

Solution

verified

Verified by Toppr

In ΔABD and ΔACD

AB=AC (given)

∠BAD=∠CAD

As AD is bisector of ∠A and AD=AD

ΔDAB=ΔDAC (by SAS congruency rule)

∠ADB=∠ADC (by c.p.c.t)

∠ADB=∠ADC=90

0

and BD=DC

In ΔABD,

AD

2

+BD

2

=AB

2

…(i)

AD

2

+DC

2

=AC

2

…(ii)

Adding (i) and (ii), we get

2AD

2

+BD

2

+DC

2

=AB

2

+AC

2

2AD

2

+BD

2

+DC

2

=BC

2

2AD

2

+2BD

2

=BC

2

2(AD

2

+BD

2

)=BC

2

2[AD

2

+(

2

1

BC)

2

]=BC

2

2[AD

2

+

4

1

BC

2

]=BC

2

2AD

2

+

2

1

BC

2

=BC

2

2AD

2

=BC

2

−

2

1

BC

2

=

2

BC

2

4AD

2

=BC

2

2AD=BC

i.e. BC=2AD

Hence proved.