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

Answer:

Proved

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!

Subscribe to EduCute for more knowledge [ Only if you want to ]

Answer 2

$\texttt{\textsf{\large{\underline{Correct Question}:}}}$

• ∆ABC is an isosceles triangle whose ∠C is a right angle. If D is the mid point on AB, prove that - AD² + BD² = 2CD²

$\texttt{\textsf{\large{\underline{Proof}:}}}$

Given Info:

1. ∆ABC is an isosceles triangle whose ∠C is a right angle.
2. D is the midpoint on AB. Therefore - AD = BD.

From ∆ABC, we get:

→ AB² = AC² + BC² [By Pythagoras Theorem]

Now consider ∆ADC and ∆BCD –

1. AC = BC (Given)

2. CD = CD (Common)

3. AD = BD (As D is the midpoint on AB)

Therefore:

→ ∆ADC ≅ ∆BCD [SSS rule of congruency]

Therefore:

→ ∠ADC = ∠BDC [C.P.C.T]

Also:

→ ∠ADC + ∠BDC = 180° [AB is a straight line]

Therefore:

→ ∠ADC = ∠BDC = 90°

★ So, ∆ADC and ∆BCD are right angled triangle.

From ∆BCD:

→ BC² = BD² + CD² – (i) [By Pythagoras Theorem]

From ∆ADC:

→ AC² = AD² + CD² – (ii) [By Pythagoras Theorem]

Adding (i) and (ii), we get:

→ AC² + BC² = AD² + BD² + 2CD²

→ AB² = AD² + BD² + 2CD²

As AB = 2AD:

→ (2AD)² = AD² + BD² + 2CD²

→ 4AD² - AD² - BD² = 2CD²

→ 3AD² - BD² = 2CD²

→ AD² + 2AD² - BD² = 2CD²

→ AD² + 2BD² - BD² = 2CD² [AD = BD]

→ AD² + BD² = 2CD²

Hence Proved!

$\texttt{\textsf{\large{\underline{Know More}:}}}$

Pythagoras Theorem: It states that in a right angled triangle, the square of the largest side is equal to the sum of the squares of the other two sides.