Q. 11
ABC is and isosceles triangle whose ∠C is right angle. If D is mid point of AB, then let us prove that, AD2 + DB2 = 2CD2
Answer :
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!
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