Question

An equilateral triangle BAP is drawn inside a square ABCD, then the value of angle DPC=

Answer 1

Given Δ ABP is an equilateral triangle.

Then, AP = BP (Same side of the triangle)

and AD = BC (Same side of square)

and, ∠ DAP = ∠ DAB - ∠ PAB = 90° - 60° = 30°

Similarly,

∠ BPC = ∠ ABC - ∠ ABP = 90° - 60° = 30°

∴ ∠ DAP = ∠ BPC

∴ Δ APD is congruent to Δ BPC ( SAS proved)

In Δ APD

AP = AD II as AP = AB (equilateral triangle)

We know that ∠ DAP = 30°

∴ ∠ APD = (180° - 30°)/2 (Δ APD is an isosceles triangle and ∠ APD is on of the base angles.)

= 150°/2 = 75°  

= ∠ APD = 75°

Similarly, ∠ BPC = 75°

Therefore, ∠ DPC = 360° - (75°+75°+60°)

= ∠ DPC = 150°

Now in Δ PDC  

PD = PC as Δ APD is congruent to Δ BPC

∴ Δ PDC is an isosceles triangle

And ∠ PDC = ∠ PCD = (180° - 150°)/2  

or ∠ PDC = ∠ PCD = 15°

∠ DPC = 150°; ∠ PDC = 15° and ∠ PCD = 15°

Answer 2

Answer :

Given Δ ABP is an equilateral triangle.

Then, AP = BP (Same side of the triangle)

and AD = BC (Same side of square)

and, ∠ DAP = ∠ DAB - ∠ PAB = 90° - 60° = 30°

Similarly,

∠ BPC = ∠ ABC - ∠ ABP = 90° - 60° = 30°

∴ ∠ DAP = ∠ BPC

∴ Δ APD is congruent to Δ BPC ( SAS proved)

In Δ APD

AP = AD II as AP = AB (equilateral triangle)

We know that ∠ DAP = 30°

∴ ∠ APD = (180° - 30°)/2 (Δ APD is an isosceles triangle and ∠ APD is on of the base angles.)

= 150°/2 = 75°  

= ∠ APD = 75°

Similarly, ∠ BPC = 75°