Answer:
See Below.
Step-by-step explanation:
We are given that ΔAPB and ΔAQC are equilateral triangles.
And we want to prove that PC = BQ.
Since ΔAPB and ΔAQC are equilateral triangles, this means that:

Likewise:

Since they all measure 60°.
Note that ∠PAC is the addition of the angles ∠PAB and ∠BAC. So:

Likewise:

Since ∠QAC ≅ ∠PAB:

And by substitution:

Thus:

Then by SAS Congruence:

And by CPCTC:

Answer:
tough question guy needed more points to answer
Step-by-step explanation:
<h2>MARK AS BRAINLIEST </h2>
Answer:
2fg
Step-by-step explanation:
Answer: The tree is 4 feet tall.
Step-by-step explanation: 5/2=x/10
2 x 10=20
20/5=4