Question

The perpendicular from A on
side BC of a ABC intersects
BC at D such that DB = 3 CD
(see figure).
Prove that
2AB2 = 2AC2 + BC?
CBSE 201
no
​

Answer 1

Hi mate here is the answer:-✍️✍️

Question:

The perpendicular from A on side BC of aΔABC intersects BC at D such that DB = 3 CD(see Fig. 6.55). Prove that 2AB2 = 2AC2 + BC2

Answer:

We have two ∆ are ∆ACD and ∆ ABD

By Pythagoras theorem in ∆ACD---

AC2 = AD2 + DC2

AC2 = AD2 + DC2AD2 = AC2 – DC2 ......................eq(i)

Applying Pythagoras theorem in ΔABD, we obtainAB2 = AD2 + DB2AD2 = AB2 – DB2 ........................eq(ii)

Now we can see from equation i and equation ii that LHS is same. Thus,

From (i) and (ii), we get

AC2 – DC2 = AB2 – DB2 (iii)

It is given that 3DC = DBTherefore,

DC + DB = BC

DC + DB = BCDC + 3DC = BC

DC + DB = BCDC + 3DC = BC4 DC = BC ......................eq(iv)

and also,

DC = DB/3

putting this in eq (iii)

DB=3BC/4

So, DC=BC/4

putting these values in eqn. (3)

AC^2-(BC)^2/4 =AB^2- 3(BC)^2/4

then step by step.......

16AB^2-16AC^2=8BC^2

2AB^2=2BC^2+AC

Hope it helps you ❣️☑️☑️☑️

Answer 2

Answer:

See the attachment. Hope it helps uh!