Question

ABC is an isosceles triangle whose ∠C is right angle. If D is any point on AB, then let us prove that, AD² + DB² = 2CD². ​

Answer 1

Step-by-step explanation:

Given:

ABC is an isosceles triangle. So, AC = BC

∠C = right angle

To prove:-

AD2 + DB2 = 2CD2

Proof:

By Pythagoras Theroem,

In ΔABC, We have

AB2 = AC2 + BC2 [1]

In ΔADC and ΔADB, we have

AC = BC [Given]

CD = CD [Common]

AD = DB [D is the mid point]

⇒ ΔADC ≅ ΔADB [By SSS Congruency Criterion]

⇒ ∠ADC = ∠ADB [Corresponding parts of congruent triangles are equal]

Also, ∠ADC + ∠ADB = 180° [Linear Pair]

⇒ ∠ADC + ∠ADC = 180°

⇒ ∠ADC = ∠ADB = 90°

Hence, ADC and ADB are right triangles, ∴ By Pythagoras theorem

AC2 = CD2 + AD2 [2]

BC2 = CD2 + BD2 [3]

Adding [2] and [3], we have

AC2 + BC2 = AD2 + BD2 + 2CD2

⇒ AB2 = AD2 + BD2 + 2CD2

⇒ (2AD)2 = AD2 + AD2 + 2CD2

[∵ D is mid-point of AB, AD = BD and AB = 2AD = 2BD]

⇒ 4AD2 = 2AD2 + 2CD2

⇒ 2AD2 = 2CD2

⇒ AD2 + BD2 = 2CD2 [∵ AD = BD]

Hence, Proved!