In the given figure, ABCD is a square and △PAB is an equilateral triangle. (i) Prove that △APD ≅ △BPC. (ii) Show that ∠DPC = 15°.

(i) We know that, Each interior angle in a square is 90° and each interior angle in an equilateral triangle is 60°. From figure, ⇒ ∠DAP = ∠DAB + ∠BAP ⇒ ∠DAP = 90° + 60° = 150° ⇒ ∠CBP = ∠CBA + ∠ABP ⇒ ∠CBP = 90° + 60° = 150° In △APD and △BPC, ⇒ ∠DAP = ∠CBP [Each equal to 150°] ⇒ AD = BC [Sides of a square] ⇒ AP = BP [Sides of an equilateral triangle] ∴ △APD ≅ △BPC (By S.A.S axiom) Hence, △APD ≅ △BPC. (ii) ABCD is a square. ∴ AB = AD = DC = BC APB is an equilateral triangle. ∴ AP = PB = AB So, we get : AP = AD and PB = BC ∴ △APD and △BPC are isosceles triangle. We know that, Angles opposite to equal sides are equal. ∴ ∠APD = ∠ADP = x (let) and ∠BPC = ∠BCP = y (let) In △APD, By angle sum property of triangle, ⇒ ∠APD + ∠ADP + ∠DAP = 180° ⇒ x + x + 150° = 180° ⇒ 2x = 180° - 150° ⇒ 2x = 30° ⇒ x = ⇒ x = 15°. In △BPC, By angle sum property of triangle, ⇒ ∠BPC + ∠BCP + ∠PBC = 180° ⇒ y + y + 150° = 180° ⇒ 2y = 180° - 150° ⇒ 2y = 30° ⇒ y = ⇒ y = 15° From figure, ⇒ ∠PDC = ∠ADC - ∠ADP = 90° - 15° = 75° ⇒ ∠PCD = ∠BCD - ∠BCP = 90° - 15° = 75° In △DPC, By angle sum property of triangle, ⇒ ∠DPC + ∠PDC + ∠PCD = 180° ⇒ ∠DPC + 75° + 75° = 180° ⇒ ∠DPC + 150° = 180° ⇒ ∠DPC = 180° - 150° ⇒ ∠DPC = 30°. Hence, proved that ∠DPC = 30°.

In the given figure, ∠ABD = ∠EBC, BD = BC and ∠ACB = ∠EDB. Prove that AB = BE.

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In the given figure, AY ⊥ ZY nd BY ⊥ XY such that AY = ZY and BY = XY. Prove that AB = ZX.

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In the given figure, in △ABC, ∠B = 90°. If ABPQ and ACRS are squares, prove that:
(i) △ACQ ≅ △ABS
(ii) CQ = BS.

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Squares ABPQ and ADRS are drawn on the sides AB and AD of a parallelogram ABCD. Prove that:
(i) ∠SAQ = ∠ABC
(ii) SQ = AC.

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Mathematics

In the given figure, ABCD is a square and △PAB is an equilateral triangle.
(i) Prove that △APD ≅ △BPC.
(ii) Show that ∠DPC = 15°.

Triangles

Related Questions

In the given figure, ∠ABD = ∠EBC, BD = BC and ∠ACB = ∠EDB. Prove that AB = BE.

In the given figure, AY ⊥ ZY nd BY ⊥ XY such that AY = ZY and BY = XY. Prove that AB = ZX.

In the given figure, in △ABC, ∠B = 90°. If ABPQ and ACRS are squares, prove that:
(i) △ACQ ≅ △ABS
(ii) CQ = BS.

Squares ABPQ and ADRS are drawn on the sides AB and AD of a parallelogram ABCD. Prove that:
(i) ∠SAQ = ∠ABC
(ii) SQ = AC.

Mathematics

In the given figure, ABCD is a square and △PAB is an equilateral triangle.(i) Prove that △APD ≅ △BPC.(ii) Show that ∠DPC = 15°.

Triangles

Related Questions

In the given figure, ∠ABD = ∠EBC, BD = BC and ∠ACB = ∠EDB. Prove that AB = BE.

In the given figure, AY ⊥ ZY nd BY ⊥ XY such that AY = ZY and BY = XY. Prove that AB = ZX.

In the given figure, in △ABC, ∠B = 90°. If ABPQ and ACRS are squares, prove that:(i) △ACQ ≅ △ABS(ii) CQ = BS.

Squares ABPQ and ADRS are drawn on the sides AB and AD of a parallelogram ABCD. Prove that:(i) ∠SAQ = ∠ABC(ii) SQ = AC.

In the given figure, ABCD is a square and △PAB is an equilateral triangle.
(i) Prove that △APD ≅ △BPC.
(ii) Show that ∠DPC = 15°.

In the given figure, in △ABC, ∠B = 90°. If ABPQ and ACRS are squares, prove that:
(i) △ACQ ≅ △ABS
(ii) CQ = BS.

Squares ABPQ and ADRS are drawn on the sides AB and AD of a parallelogram ABCD. Prove that:
(i) ∠SAQ = ∠ABC
(ii) SQ = AC.