Mathematics
In a quadrilateral ABCD, ∠A + ∠D = 90°, prove that : AC2 + BD2 = AD2 + BC2.
Answer
Produce AB and DC such that they meet at point E.
By formula,
By pythagoras theorem,
⇒ (Hypotenuse)2 = (Perpendicular)2 + Base2
In △ AED,
⇒ ∠A + ∠D + ∠E = 180°
⇒ 90° + ∠E = 180°
⇒ ∠E = 180° - 90° = 90°.
By pythagoras theorem,
⇒ AD2 = AE2 + DE2 ………(1)
In △ BEC,
By pythagoras theorem,
⇒ BC2 = BE2 + CE2 ………(2)
In △ AEC,
By pythagoras theorem,
⇒ AC2 = AE2 + CE2 ………(3)
In △ BED,
By pythagoras theorem,
⇒ BD2 = BE2 + DE2 ………(4)
Adding equations (1) and (2), we get :
⇒ AD2 + BC2 = AE2 + DE2 + BE2 + CE2
⇒ AD2 + BC2 = (AE2 + CE2) + (BE2 + DE2)
⇒ AD2 + BC2 = AC2 + BD2.
Hence, proved that AC2 + BD2 = AD2 + BC2.
Related Questions
In triangle ABC, angle A = 90°, CA = AB and D is a point on AB produced. Prove that :
DC2 - BD2 = 2AB.AD.
In triangle ABC, AB = AC and BD is perpendicular to AC. Prove that :
BD2 - CD2 = 2CD × AD
In the following figure, AD is perpendicular to BC and D divides BC in the ratio 1 : 3.
Prove that : 2AC2 = 2AB2 + BC2.
In the given figure, AB = 16 cm, BC = 12 cm and CA = 6 cm; find the length of CD.
