In triangle ABC AB is equal to AC and D is a point on BC prove that AB square minus A D square is equal to BD into CD
Answers
Step-by-step explanation:
Prove that : In ∆ABE and ∆ACE, we have
AB = AC [given]
AE = AE [common]
and ∠AEB = ∠AEC [90°]
Therefore, by using RH congruent condition
∆ABE ~ ∆ACE
⇒ BE = CE
In right triangle ABE.
AB2 = AE2 + BE2 ...(i)
[Using Pythagoras theorem]
In right triangle ADE,
AD2 = AE2 + DE2
[Using Pythagoras theorem]
Subtracting (ii) from (i), we get
AB2 - AD2 = (AE2 + BE2) - (AE2 + DE2)
AB2 - AD2 = AE2 + BE2 - AE2 - DE2
⇒ AB2 - AD2 = BE2 - DE2
⇒ AB2 - AD2 (BE + DE) (BE - DE)
But BE = CE [Proved above]
⇒ AB2 - AD2 = (CE + DE) (BE - DE)
= CD.BD
⇒ AB2 - AD2 = BD×CD
Hence Proved...
Answer:
sol: in ∆ ABC side AB is equal to side AC
line D is the midpoint of side BC
In ∆ABD,
by phythagor theorm
AB sq = AD sq + BD sq.......(1)
In ∆ACD,
by phythagors theorm
AC sq = AD sq + CD sq.......(2)
from (1)and (2)
AB sq - AC sq = BD sq - CD sq
Step-by-step explanation:
sorry I am getting this ans