Question

abcd is a square o. the side dc on equilateral triangle is constructed. prove that ae=be and angle dac=15 degree

Answer 1

Given : ABCD is a square and EDC is an equilateral triangle. AD = BC, DE = CE To Prove : i) AE = BE ii) ∠DAE = 15°

Construction : Join A to E and B to E.

Proof : I)In ΔADE and ΔBCE, AD = BC (given)

∠ADE = ∠BCE (90° + 60°)

DE = CE (given)

Therefore, ΔADE is congruent to ΔBCE ( SAS rule) AE = BE (CPCTC)

ii) ∠DAE + ∠ADE + ∠DEA
= 180° 150° + ∠DAE + ∠DEA
= 180° ∠DAE + ∠DEA
= 180° -150° 2 ∠DAE
= 30° ∠DAE = 15° Hence Proved.



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Answer 2

Answer:

Given : ABCD is a square and EDC is an equilateral triangle. AD = BC, DE = CE To Prove : i) AE = BE ii) ∠DAE = 15°

Construction : Join A to E and B to E.

Proof : I)In ΔADE and ΔBCE, AD = BC (given)

∠ADE = ∠BCE (90° + 60°)

DE = CE (given)

Therefore, ΔADE is congruent to ΔBCE ( SAS rule) AE = BE (CPCTC)

ii) ∠DAE + ∠ADE + ∠DEA

= 180° 150° + ∠DAE + ∠DEA

= 180° ∠DAE + ∠DEA

= 180° -150° 2 ∠DAE

= 30° ∠DAE = 15° Hence Proved.

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