Question

Please answer
Don't spam please

​

Answer 1

Answer:

Construction: From the point D perpendicular DE and DF are drawn on BC and CA.

Proof: Now ∠DFC = 1 right angle

and ∠DEC = 1 right angle

and ∠DEC = 1 right angle

∴ In quadrileter CEDF,

∠FCE = ∠DFC = ∠DEC = right angle.

∴ CEDF is a rectangle.

∴ FD = CE and CF = ED

In rectangle CEDF, CF || ED

i.e CA || ED and ADB is the interceptor.

∴ Similar ∠CAD = similar ∠EDB

∴ ∠EDB = ∠DBE

[ $\because$ ∠CAB = ∠ABC because ∆ABC is an isosceles triangle]

$\because$ In ∆BDE, ∆CDE, ∆BDE

∴ DE = BE

∴ CF = D E = BE.

Now, ∆ADF, ∆BDE and ∆CDE

∴ AD² = AF² + FD²,

DB² = BE² + DE²,

CD² = CE² + DE²

Now, AD² + DB²,

➻ AF² + FD² + BE² +DE²

➻ CE² + CE² + DE² + DE²

➻ 2CE² + 2DE²

➻ 2(CE² + DE²)

➻ 2CD²

∴ AD² + DB² = 2CD²

(PROVED)

Answer 2

EXPLANATION.

ABC is an isosceles triangle,

∠C is a right angled triangle.

D is any point on AB.

As we know that,

Pythagoras theorem,

⇒ H² = P² + B².

Using the formula in this question, we get.

In ΔABC,

⇒ (AB)² = (AC)² + (BC)². ⇒(1).

In ΔADC = BDC.

⇒ AC = BC.

⇒ CD = CD [ Common].

⇒ AD = DB.

⇒ D is a mid point.

⇒ ΔADC ≅ ΔBDC [ S.S.S Criteria ].

⇒ ∠ADC = ∠BDC [ Corresponding pain of conjugate triangle are equal ].

⇒ ∠ADC + ∠BDC = 180°.

⇒ ∠ADC + ∠ADC = 180°.

⇒ 2∠ADC = 180°.

⇒ ∠ADC = 90°.

⇒ ∠BDC + ∠BDC = 180°.

⇒ 2∠BDC = 180°.

⇒ ∠BDC = 90°.

⇒ ∠ADC = ∠BDC = 90°.

In ΔADC,

⇒ (AC)² = (CD)² + (AD)². ⇒(2).

In ΔBDC,

⇒ (BC)² = (CD)² + (BD)². ⇒(3).

From equation (2) & (3), we get.

Adding equation (2) & (3), we get.

⇒ (AC)² + (BC)² = (CD)² + (AD)² + (CD)² + (BD)².

⇒ (AC)² + (BC)² = (AD)² + (BD)² + (2CD)².

As we know that,

⇒ AD = DB.

⇒ AD = AD.

⇒ AB = 2AD.

⇒ (2AD)² = (AD)² + (BD)² + (2CD)².

⇒ (4AD)² = (AD)² + (BD)² + (2CD)².

⇒ (4AD)² = (AD)² + (AD)² + (2CD)².

⇒ (4AD)² = (2AD)² + (2CD)².

⇒ (2AD)² = (2CD)².

⇒ (AD²) + (AD²) = (2CD)².

⇒ (AD)² + (BD)² = (2CD)².

HENCE PROVED.