Question

In a square ABCD, point P is chosen inside ABCD and point Q outside ABCD such that APB and BQC are congruent isosceles triangles with angle APB = angle BQC = 80 degrees. T is a point where BC and PQ meet. Find the size of angle BTQ

Answer 1

Since BQ=BP and ∡QBP=90∘,

We obtain ∡BPT=45∘

And

∡BTQ=∡BPT+∡PBT=45∘+40∘=85∘

Please leave a comment below if you have any problems.

Answer 2

Answer:

∠BTQ is 85°

Step-by-step explanation:

Given In a square ABCD, point P is chosen inside ABCD and point Q outside ABCD such that APB and BQC are congruent isosceles triangles with angle APB = angle BQC = 80 degrees. T is a point where BC and PQ meet.we have to find the angle ∠BTQ=∠3

In ∠1=∠9=80°

In ΔBQC, ∠1+∠4+∠5=180°

⇒ 80°+2∠4=180°       (∵BQC is isosceles triangle)

⇒ 2∠4=100°  ⇒  ∠4=50°

In ΔABP, ∠8+∠9+∠10=180°

⇒ 80°+2∠8=180°       (∵ABP is isosceles triangle)

⇒ 2∠8=100°  ⇒  ∠8=50°

∵ APB and BQC are congruent isosceles triangles

By CPCT, AP=BQ but AP=PB ⇒BQ=PB

Therefore, PBQ is isosceles triangle, ∠2=∠11

∠6=∠7+∠8 ⇒ ∠7=∠6-∠8=90°-50°=40°

In ΔPBQ, ∠2+∠4+∠7+∠11=180°     (angle sum property)

⇒∠2+40°+50°+∠2=180°⇒ 2∠2=90°  ⇒ ∠2=45°

In ΔBTQ, ∠2+∠3+∠4=180°

⇒ 45°+∠3+50°=180°

⇒ ∠3=180°-95°=85°