Advertisements
Advertisements
प्रश्न
ABCD is a square E, F, G and H are points on AB, BC, CD and DA respectively, such that AE = BF = CG = DH. Prove that EFGH is a square.
Advertisements
उत्तर १

We have
AE = BF = CG = DH = x (say)
∴ BE = CF = DG = AH = y (say)
In Δ ' s AEH and BEF , we have
AE = BF
`∠`A = `∠`B
And AH = BE
So, by SAS configuration criterion, we have
ΔAEH ≅ ΔBFE
⇒ `∠`1 = `∠`2 and `∠`3 = `∠`4
But `∠`1+ `∠`3 = 90° and `∠`2 + `∠`4 = 90°
⇒ `∠`1+ `∠`3+ `∠`2 + `∠`4 = 90° + 90°
⇒ `∠`1+ `∠`4 + `∠`1+ `∠`4 = 180°
⇒ 2 (`∠`1+ `∠`4) = 180°
⇒ `∠`1+ `∠`4 = 90°
HEF = 90°
Similarly we have `∠`F = `∠`G = `∠`H = 90°
Hence, EFGH is a square
उत्तर २

We have
AE = BF = CG = DH = x (say)
∴ BE = CF = DG = AH = y (say)
In Δ ' s AEH and BEF , we have
AE = BF
`∠`A = `∠`B
And AH = BE
So, by SAS configuration criterion, we have
ΔAEH ≅ ΔBFE
⇒ `∠`1 = `∠`2 and `∠`3 = `∠`4
But `∠`1+ `∠`3 = 90° and `∠`2 + `∠`4 = 90°
⇒ `∠`1+ `∠`3+ `∠`2 + `∠`4 = 90° + 90°
⇒ `∠`1+ `∠`4 + `∠`1+ `∠`4 = 180°
⇒ 2 (`∠`1+ `∠`4) = 180°
⇒ `∠`1+ `∠`4 = 90°
HEF = 90°
Similarly we have `∠`F = `∠`G = `∠`H = 90°
Hence, EFGH is a square
APPEARS IN
संबंधित प्रश्न
In a ∆ABC, D, E and F are, respectively, the mid-points of BC, CA and AB. If the lengths of side AB, BC and CA are 7 cm, 8 cm and 9 cm, respectively, find the perimeter of ∆DEF.
In Fig. below, triangle ABC is right-angled at B. Given that AB = 9 cm, AC = 15 cm and D,
E are the mid-points of the sides AB and AC respectively, calculate
(i) The length of BC (ii) The area of ΔADE.

In Fig. below, M, N and P are the mid-points of AB, AC and BC respectively. If MN = 3 cm, NP = 3.5 cm and MP = 2.5 cm, calculate BC, AB and AC.

Show that the line segments joining the mid-points of the opposite sides of a quadrilateral
bisect each other.
In the given figure, ΔABC is an equilateral traingle. Points F, D and E are midpoints of side AB, side BC, side AC respectively. Show that ΔFED is an equilateral traingle.

In the adjacent figure, `square`ABCD is a trapezium AB || DC. Points M and N are midpoints of diagonal AC and DB respectively then prove that MN || AB.

In triangle ABC ; D and E are mid-points of the sides AB and AC respectively. Through E, a straight line is drawn parallel to AB to meet BC at F.
Prove that BDEF is a parallelogram. If AB = 16 cm, AC = 12 cm and BC = 18 cm,
find the perimeter of the parallelogram BDEF.
In triangle ABC; M is mid-point of AB, N is mid-point of AC and D is any point in base BC. Use the intercept Theorem to show that MN bisects AD.
E is the mid-point of the side AD of the trapezium ABCD with AB || DC. A line through E drawn parallel to AB intersect BC at F. Show that F is the mid-point of BC. [Hint: Join AC]
P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD in which AC = BD. Prove that PQRS is a rhombus.
