Question

In the following diagrams, ABCD is a square and APB is an equilateral triangle .in each case, prove that triangle APD is congruent triangle BPC and find the angles of triangle DPC.

Answer 1

Solution : part 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)
2nd part
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.

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)

2nd part

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°