state and prove :-
1.Basic proportionality.....
2. Converse Basic proportionality....
3.Phythagoras theorem....
4.Converse Pythagoras theorem.....
(with the help of diagrams)
Answers
Given In ∆ABC , DE || BC and intersects AB in D and AC in E.
To Prove : AD/DB = AE/EC
Proof : 1) EF ┴ BA
3)Area(∆ADE) = (AD .EF)/2
4)Area(∆DBE) =(DB.EF)/2
5)(Area(∆ADE))/(Area(∆DBE)) = AD/DB
6) (Area(∆ADE))/(Area(∆DEC)) = AE/EC
7) ∆DBE ~∆DEC
8) Area(∆DBE)=area(∆DEC)
9) AD/DB =AE/EC
Hence Proved..
Converse of BPT Theorem..
Given
In ΔABC, D and E are the two points of AB and AC respectively,
such that .
To Prove
DE || BC
Proof
2) AD / DB = AF / FC
3) AD / DB = AE /EC
4) AF / FC = AE / EC
5) (AF/FC) + 1 = (AE/EC) + 1
6) (AF + FC )/FC = (AE + EC)/EC
7) AC /FC = AC / EC
8) FC = EC
Hence Proved...
Pythagoras Theorem..
Given : A right ΔABC right angled at B
To prove : AC2 = AB2 + BC2
Construction : Draw AD ⊥ AC
Proof : ΔABD and ΔABC
∠ADB = ∠ABC = 90°
∠BAD = ∠BAC (common)
∴ ΔADB ∼ ΔABC (by AA similarly criterion)
AD/AB=AB/AC
⇒ AD × AC = AB² ...... (1)
Now In ΔBDC and ΔABC
∠BDC = ∠ABC = 90°
∠BCD = ∠BCA (common)
∴ ΔBDC ∼ ΔABC (by AA similarly criterion)
CD/BC = BC/AC
⇒ CD × AC = BC² ........ (2)
Adding (1) and (2) we get
AB2 + BC2 = AD × AC + CD × AC
= AC (AD + CD)
= AC × AC = AC²
∴ AC² = AB² + BC²
Hence Proved....
Converse of Pythagoras Theorem
GIVEN:− ABC is a △ in which AC² = AB² + BC²
To Prove:− ∠B = 90°
Construction− Draw a △ PQR right angled at Q, such that QR = BC and PQ = AB
Proof:-In △ ABC,AC2 = AB2 + BC2 given ..........1
Now, PQ = AB By construction
QR = BC
∴ From 1, AC² = PQ² + QR² ..........2
Now, from △ PQR,
PR² = PQ² + QR² Pythagoras theorem
∴ from 2, we get, AC² = PR²
⇒ AC = PR
Now, In △ ABC and △ PQR,
AB = PQ By construction
BC = QR By construction
AC = PR Proved above
⇒ ABC ≅ △ PQR by SSS
⇒ ∠ B = ∠ Q by CPCT
⇒ ∠ B = 90°
Hence Proved..
I proved All Hope This Help:
I reffered Google for Images
