❥Prove that the sum of squares of sides of a parallelogram is equal to the sum of the squares of it's diagonals.
Let ABC be a right-angled triangle with ∠B = 90◦.Let I be the incentre of ABC. Draw a line perpendicular to AI at I. Let it intersect the line CB at D. Prove that CI is perpendicular to AD and prove that ID =Sqrt(b(b − a)) ,where BC = a and CA = b.
Even though it is a question from RMO, doesn't mean it will be difficult , it can be solved elegantly with using only very basic tools like Pythagoras theorem, similarity ,incenter relation with triangle etc.
────┈┈┈┄┄╌╌╌╌┄┄┈┈┈────
⚠no spam⚠
quality answer⚡
Answers
Answer:
Step-by-step explanation:
Hey nidhi sis here is you answer:
___________________________________
Question :Prove that the sum of squares of sides of a parallelogram is equal to the sum of the squares of it's diagonals.
Answer :In parallelogram ABCD, AB = CD, BC = AD
Draw perpendiculars from C and D on AB as shown.
In right angled ΔAEC, AC2 = AE2 + CE2 [By Pythagoras theorem]
⇒ AC2 = (AB + BE)2 + CE2
⇒ AC2 = AB2 + BE2 + 2 AB × BE + CE2 → (1)
From the figure CD = EF (Since CDFE is a rectangle)
But CD= AB
⇒ AB = CD = EF
Also CE = DF (Distance between two parallel lines)
ΔAFD ≅ ΔBEC (RHS congruence rule)
⇒ AF = BE
Consider right angled ΔDFB
BD2 = BF2 + DF2 [By Pythagoras theorem]
= (EF – BE)2 + CE2 [Since DF = CE]
= (AB – BE)2 + CE2 [Since EF = AB]
⇒ BD2 = AB2 + BE2 – 2 AB × BE + CE2 → (2)
Add (1) and (2), we get
AC2 + BD2 = (AB2 + BE2 + 2 AB × BE + CE2) + (AB2 + BE2 – 2 AB × BE + CE2)
= 2AB2 + 2BE2 + 2CE2
AC2 + BD2 = 2AB2 + 2(BE2 + CE2) → (3)
From right angled ΔBEC, BC2 = BE2 + CE2 [By Pythagoras theorem]
Hence equation (3) becomes,
AC2 + BD2 = 2AB2 + 2BC2
= AB2 + AB2 + BC2 + BC2
= AB2 + CD2 + BC2 + AD2
∴ AC2 + BD2 = AB2 + BC2 + CD2 + AD2
Thus the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.
___________________________________
Question: Let ABC be a right-angled triangle with ∠B = 90◦.Let I be the in centre of ABC. Draw a line perpendicular to AI at I. Let it intersect the line CB at D. Prove that CI is perpendicular to AD and prove that ID =√(b(b − a)) ,where BC = a and CA = b.
Answer:
First observe that ADBI is a cyclic quadrilateral since ∠AID=∠ABD=90°.
Hence ∠ADI=∠ABI=45°. But we also have:
∠ADB = ∠ADI + ∠IDB = 45◦ + ∠IAB
= ∠DAI + ∠IAC = ∠DAC.
Therefore CDA is an isosceles triangle with CD =
CA. Since CI bisects ∠C it follows that CI ⊥ AD.
This shows that DB = CA − CB = b − a. Therefore
AD²=c²+(b-a)²=c²+b²+a²-2ba=2b(b-a),
But then 2ID²=AD²-2b(b-a)and this gives ID =√b(b-a).
___________________________________
Hope it helps you sis :
────┈┈┈┄┄╌╌╌╌┄┄┈┈┈────
Prove that the sum of squares of sides of a parallelogram is equal to the sum of the squares of it's diagonals.
Given: ABCD is a parallelogram in which AC and BD are diagonals which intersect each other at 0.
To Prove: AB²+ BC² + CD² + DA² = AC² + BD²
Construction: Draw AX ⊥ CD and BY ⊥ CD extended to Y.
Proof:
Let us consider ∆AXC.
AC² = AX² + CX² | (1)
Consider the ∆BYD.
BD² = BY² + DY² | (2)
From the figure, DY = CD + CY.
Thus in equation (2),
BD² + BY² + DY² = BY² + (CD + CY)²
∵ (a + b)² = a² + b² + 2ab
∴ BD² = (BY² + CY²) + CD² +2CD.CY | (3)
Now using Pythagoras theorem in ∆BYC,
We get BC² = BY² + CY² | (4)
Substituting the value of equation (4) in (3)
We get BD² = (BY² + CY²) + CD² +2CD.CY | (5)
Now let us consider equation (1)
AC² = AX² + CX²
From figure, we can say that CD = CX + XD
∴ CX = CD - DX
∵ (a + b)² = a² + b² + 2ab
∴AC² = AX² + (CD - CX)²
AC² = AX² + CD² + CX² - 2CD.DX
AC² = (AX² + CX²) +CD² - 2CD.DX | (6)
Now using Pythagoras theorem in ∆AXD,
We get AD² = AX² + DX² | (7)
Substituting the value of equation (7) in (5)
We get AC² = AD² + CD² -2CD.DX
As opposite sides of parallelogram are equal.
We get, AB = CD
AC² = AD² + AB² -2CD.DX | (8)
Now by adding equations (5) and (8)
We get, BD² + AC² = BC² + CD² + AD² + AB² + 2CD.CY - 2CD.DX
∴BD² + AC² = AB² + BC² + CD² + AD² + 2CD(CY-DX) | (9)
From the figure, let us consider ∆AXD and ∆BYC. ∠AXD = ∠BYC | Both are 90° by construction.
AX = BY | The perpendicular drawn between two parallel lines are equal.
AD = BC | Opposite sides of a parallelogram are equal.
∴∆AXD = ∆BYC | By SAS Cong. Rule
Now CY = DX | By CPCT
Putting CY = DX in eq (9)
We get,
BD² + AC² = AB² + BC² + CD² + AD² + 2CD(DX-DX)
BD² + AC² = AB² + BC² + CD² + AD² + 2CD x 0
BD² + AC² = AB² + BC² + CD² + AD²
Hence proved
╍╍╍╍╍╍╍╍╍╍╍╍╍╍╍╍╍╍╍╍╍╍╍╍╍╍╍╍╍╍╍╍╍╍╍╍╍╍╍╍╍╍╍
Let ABC be a right-angled triangle with ∠B = 90◦.Let I be the incentre of ABC. Draw a line perpendicular to AI at I. Let it intersect the line CB at D. Prove that CI is perpendicular to AD and prove that ID =Sqrt(b(b − a)) ,where BC = a and CA = b.
We can observe that ADBI is a cyclic quadrilateral since ∠AID = ∠ABD = 90°.
Hence ∠ADI = ∠ABI = 45° and hence ∠DAI = 45°.
We also have ∠ADB = ∠ADI + ∠IDB | (1)
Substituting ∠ADI = ∠ABI = 45° in eq. 1, we get
∠ADB = 45° + ∠IDB
= ∠DAI + ∠IAC = ∠DAC
Therefore CDA is an isosceles triangle with CD = CA.
Since CI bisects ∠C it follows that CI ⊥ AD.
This shows that DB = CA - CB= b - a.
∴AD² = c² + (b – a)² = c² +b² + a² – 2ba = 2b(b – a).
But because 21D² = AD² = 2b(b – a)
This gives ID =
☵☵☵☵☵☵☵☵☵☵☵☵☵☵☵☵☵☵☵☵☵☵☵☵☵☵☵☵☵☵☵☵☵☵☵☵☵
A cyclic quadrilateral is a quadrilateral for which a circle can be circumscribed so that it touches each polygon vertex. A quadrilateral that can be both inscribed and circumscribed on some pair of circles is known as a bicentric quadrilateral.
If two sides and the included angle of one triangle are equal to two sides and included angle of another triangle, then the triangles are congruent.
────┈┈┈┄┄╌╌╌╌┄┄┈┈┈────
