Prove that theta angle = 90 degrees

[petarantes]

Knowing that BQRP it is a square proves that. =90°

I'm an older student reviewing for a placement test; please check my work.
I tried to relate the theta, alpha and beta angles but all attempts generate equalities
For example 180-theta-alpha -beta - 90 + alpha+beta+theta = 90 --> 90 = 90

 

[Dr.Peterson]

Please state the problem exactly as given to you, so we can be sure what is given, and what you are to prove.

EDIT:
I have confirmed (without proof) that it is true that if BQRP is a square, then angle theta is a right angle, so that is what you are asking.

I see no proof yet; but since it involves an intersection, I would not expect that merely keeping track of angles would be enough. Would it make sense to suppose that trigonometry could be involved?

This looks like it must be a rather advanced placement test! It is not (yet) obvious to me ...

 

[petarantes]

Dr. Peterson,
The problem was transcribed correctly. I also couldn't see the solution. If we can demonstrate that alpha + beta = -90 + alpha + beta + theta the problem is solved

 

[Dr.Peterson]

Dr. Peterson,
The problem was transcribed correctly. I also couldn't see the solution. If we can demonstrate that alpha + beta = -90 + alpha + beta + theta the problem is solved

The reason I asked for the actual problem was that you omitted theta when you stated it initially:

Knowing that BQRP it is a square proves that. =90°

Then I saw that theta was mentioned in the title, which I often overlook.

May I ask again, what level of math this placement test is for? I need to know whether to expect a simple solution, or a profound one!

 

[petarantes]

The exercise was posted on a forum but does not specify the level. One tip I saw was to use Ceva's theorem.

 

[petarantes]

Follows a solution from a colleague

[MATH]\\\frac{AP}{PB} \cdot \frac{BQ}{QC} \cdot \frac {CH}{HA}=1\\ \\ PB=PR,BQ=QR\implies \frac{AP}{PR} \cdot \frac{QR}{QC} \cdot \frac {CH}{HA}=1\\ \Delta_{ABC} \sim\Delta_{APR}\sim \Delta_{RQC} \implies \frac{AB}{BC} \cdot \frac{AB}{BC} \cdot \frac {CH}{HA}=1\\ \implies \frac{AH}{CH} = \frac{AB^2}{BC^2}\\ \implies \frac{AH}{CH} + 1 = \frac{AB^2}{BC^2} +1\\ \implies \frac{AH+CH}{CH} = \frac{AB^2+BC^2}{BC^2} \\ \implies \frac{AC}{CH} = \frac{AC^2(*Pythagorean ~Theorem:\Delta_{ABC})}{BC^2}\\ \implies \frac{AC}{CH} = \frac{AC^2}{BC^2}\\ \implies \frac{\cancel{AC}}{CH} = \frac{AC\cancel{^ 2}}{BC.BC}\\ \implies \frac{AC}{BC} = \frac{BC}{CH}\\ \implies \triangle ABC \sim \triangle BHC\\ \implies\boxed{ B\hat{H}C =\theta = 90^o} [/MATH]