Question

$\huge\bold\blue{Question}$

I need the solution of the question provided in attachment .

$\huge\bold\purple{To\:Prove}$

BD = DC

$\huge\bold\red{Note}$

No spamming please !!

Answer 1

hope this helps u

given

AB=AC

AD⊥BC

to prove

BD=DC

solution

in triangle ABD and triangle ACD

AB=AC..................(i)

∠b=∠c(angle opposite to equal side are equal ).....(ii)

AD=AD as AD is common ...........(iii)

from (i)(ii)(iii)

triangle ABD ≅triangle ACD

by CPCT

BD=DC

Answer 2

Step-by-step explanation:

$\huge\bold\blue{ANSWER}$

TO PROVE

BD =DC

GIVEN

AB=DC

AD_|_BC

Therefore,

In ΔABD and ΔACD

=>AB = DC {given}

=>/_ADB =/_ADC => 90°

=>AD=AD [Common]

So,

ΔABD ~ΔACD [By ASA Rule]

=> BD=DC [ By CPCT ]

Hope It Helps ❤❤