हिंदी

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
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

 

 

shaalaa.com

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

 

 

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 13: Quadrilaterals - Exercise 13.3 [पृष्ठ ४३]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 9
अध्याय 13 Quadrilaterals
Exercise 13.3 | Q 7 | पृष्ठ ४३

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.


ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that

  1. D is the mid-point of AC
  2. MD ⊥ AC
  3. CM = MA = `1/2AB`

In below Fig, ABCD is a parallelogram in which P is the mid-point of DC and Q is a point on AC such that CQ = `1/4` AC. If PQ produced meets BC at R, prove that R is a mid-point of BC.


In triangle ABC, M is mid-point of AB and a straight line through M and parallel to BC cuts AC in N. Find the lengths of AN and MN if Bc = 7 cm and Ac = 5 cm.


In triangle ABC, AD is the median and DE, drawn parallel to side BA, meets AC at point E.
Show that BE is also a median.


In ΔABC, BE and CF are medians. P is a point on BE produced such that BE = EP and Q is a point on CF produced such that CF = FQ. Prove that: QAP is a straight line.


Prove that the figure obtained by joining the mid-points of the adjacent sides of a rectangle is a rhombus.


In ΔABC, D, E and F are the midpoints of AB, BC and AC.
If AE and DF intersect at G, and M and N are the midpoints of GB and GC respectively, prove that DMNF is a parallelogram.


In a parallelogram ABCD, E and F are the midpoints of the sides AB and CD respectively. The line segments AF and BF meet the line segments DE and CE at points G and H respectively Prove that: EGFH is a parallelogram.


In AABC, D and E are two points on the side AB such that AD = DE = EB. Through D and E, lines are drawn parallel to BC which meet the side AC at points F and G respectively. Through F and G, lines are drawn parallel to AB which meet the side BC at points M and N respectively. Prove that BM = MN = NC.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×