मराठी

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. - Mathematics

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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.

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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
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पाठ 13: Quadrilaterals - Exercise 13.3 [पृष्ठ ४३]

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आरडी शर्मा Mathematics [English] Class 9
पाठ 13 Quadrilaterals
Exercise 13.3 | Q 7 | पृष्ठ ४३

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