Mathematics
In a quadrilateral ABCD, ∠B = ∠D = 90°. Prove that : 2AC2 - BC2 = AB2 + AD2 + DC2.
Pythagoras Theorem
3 Likes
Answer
Given: ABCD is a quadrilateral where ∠B = ∠D = 90°
To prove: 2AC2 - BC2 = AB2 + AD2 + DC2
Construction: Join diagonal AC.
Proof: In Δ ABC, using Pythagoras theorem,
Hypotenuse2 = Base2 + Height2
⇒ AC2 = AB2 + BC2 ……………(1)
And, similarly in Δ ADC, using Pythagoras theorem
⇒ AC2 = AD2 + CD2 ……………(2)
Adding (1) and (2), we get
⇒ AC2 + AC2 = AB2 + BC2 + AD2 + CD2
⇒ 2AC2 = AB2 + BC2 + AD2 + CD2
⇒ 2AC2 - BC2 = AB2 + AD2 + CD2
Hence, 2AC2 - BC2 = AB2 + AD2 + CD2.
Answered By
3 Likes
Related Questions
In the following figure, straight lines l, m and n are parallel to each other and G is the mid-point of CD. Find :
(i) BG, if AD = 12 cm
(ii) CF, if GE = 4.6 cm
(iii) AB, if BC = 4.8 cm
(iv) ED, if FD = 8.8 cm
A right triangle has hypotenuse of length p cm and one side of length q cm. If p - q = 1, find the length of the third side of the triangle.
ABC is an equilateral triangle. P is a point on BC such that BP : PC = 2 : 1. Prove that : 9AP2 = 7AB2
In the following figure, ∠ABC = 90°, AB = (x + 8) cm, BC = (x + 1) cm and AC = (x + 15) cm. Find the lengths of the sides of the triangle.
