1. ## rectangle

Let ABCD be a rectangle with BC=3AB .
Let P and Q are two points on side BC with PB=PQ=QC .

Show that

< DPC +<DBC =< DQC .

2. Hello,

Originally Posted by perash
Let ABCD be a rectangle with BC=3AB .
Let P and Q are two points on side BC with PB=PQ=QC .

Show that

< DPC +<DBC =< DQC .
For that, I will use a rough method, using more algebra than geometry.

----------------------------
Because I don't remember the formula for $\arctan a+\arctan b$, I'll show it.

~~~~~~~~~~~~~~
For that I will need to find the formula (which I don't remember either) of tan(c+d) :
$\tan(c+d)=\frac{\sin(c+d)}{\cos(c+d)}$

$\tan(c+d)=\frac{\sin(c)\cos(d)+\cos(c)\sin(d)}{\co s(c)\cos(d)-\sin(c)\sin(d)}$

Let's multiply by $\frac{~~~~\frac{1}{cos(c)\cos(d)}~~~~}{~~~~\frac{1 }{cos(c)\cos(d)}~~~~}$

This gives :

$\tan(c+d)=\frac{\tfrac{\sin(c)}{\cos(c)}+\tfrac{\s in(d)}{\cos(d)}}{1-\tfrac{\sin(c)}{\cos(c)} \cdot \tfrac{\sin(d)}{\cos(d)}}$

$\boxed{\tan(c+d)=\frac{\tan(c)+\tan(d)}{1-\tan(c)\tan(d)}}$

~~~~~~~~~~~~~~

Therefore, we have :

$\tan(\arctan(a)+\arctan(b))=\frac{\tan(\arctan(a)) +\tan(\arctan(b))}{1-\tan(\arctan(a))\tan(\arctan(b))}$

$\tan(\arctan(a)+\arctan(b))=\frac{a+b}{1-ab}$

$\implies \boxed{\arctan(a)+\arctan(b)=\arctan\left(\frac{a+ b}{1-ab}\right)}$

----------------------------

Now let's go back to the problem.

So we have $\angle DPC+\angle DBC=\arctan(1/3)+\arctan(1/2)$

$\arctan(1/3)+\arctan(1/2)=\arctan\left(\frac{1/3+1/2}{1-(1/2)*(1/3)}\right)$ (formula above).

$\arctan(1/3)+\arctan(1/2)=\arctan\left(\frac{5/6}{1-1/6}\right)=\boxed{\arctan(1)}$

Therefore, $\boxed{\angle DPC+\angle DBC=\angle DQC}$

3. Originally Posted by Moo
Hello,

For that, I will use a rough method, using more algebra than geometry.

----------------------------
Because I don't remember the formula for $\arctan a+\arctan b$, I'll show it.

~~~~~~~~~~~~~~
For that I will need to find the formula (which I don't remember either) of tan(c+d) :
$\tan(c+d)=\frac{\sin(c+d)}{\cos(c+d)}$

$\tan(c+d)=\frac{\sin(c)\cos(d)+\cos(c)\sin(d)}{\co s(c)\cos(d)-\sin(c)\sin(d)}$

Let's multiply by $\frac{~~~~\frac{1}{cos(c)\cos(d)}~~~~}{~~~~\frac{1 }{cos(c)\cos(d)}~~~~}$

This gives :

$\tan(c+d)=\frac{\tfrac{\sin(c)}{\cos(c)}+\tfrac{\s in(d)}{\cos(d)}}{1-\tfrac{\sin(c)}{\cos(c)} \cdot \tfrac{\sin(d)}{\cos(d)}}$

$\boxed{\tan(c+d)=\frac{\tan(c)+\tan(d)}{1-\tan(c)\tan(d)}}$

~~~~~~~~~~~~~~

Therefore, we have :

$\tan(\arctan(a)+\arctan(b))=\frac{\tan(\arctan(a)) +\tan(\arctan(b))}{1-\tan(\arctan(a))\tan(\arctan(b))}$

$\tan(\arctan(a)+\arctan(b))=\frac{a+b}{1-ab}$

$\implies \boxed{\arctan(a)+\arctan(b)=\arctan\left(\frac{a+ b}{1-ab}\right)}$

----------------------------

Now let's go back to the problem.

So we have $\angle DPC+\angle DBC=\arctan(1/3)+\arctan(1/2)$

$\arctan(1/3)+\arctan(1/2)=\arctan\left(\frac{1/3+1/2}{1-(1/2)*(1/3)}\right)$ (formula above).

$\arctan(1/3)+\arctan(1/2)=\arctan\left(\frac{5/6}{1-1/6}\right)=\boxed{\arctan(1)}$

Therefore, $\boxed{\angle DPC+\angle DBC=\angle DQC}$

. Why go to the arctans? The tans will do.

If (angle DPC) +(angle DBC) = angle DQC
Then, tan[<DPC +<DBC) = tan(<DQC)

[tan(<DPC) +tan(<DBC)] /[1 -tan(<DPC)*tan(<DBC)] =? tan(<DQC)
[1/3 +1/2] /[1 -(1/3)(1/2)] =? 1
[5/6] /[5/6] =? 1
1 =? 1
Yes, so <DPC +<DBC = <DQC

4. Originally Posted by ticbol
. Why go to the arctans? The tans will do.
Because I imagined it this way on my sketch

Thank you for poiting that out

5. Proof using geometry

As < DQC = 45, what we need to prove is that:
< DPC + < DBC = 45
Construct another rectangle BCEF with CE = BF = AB.
R and S are points on EF such that FR = RS = SE.

Connect B and S.
triangle BQS triangle DCP are congruent, so
< SBQ = < DPC
Therefore
< DPC + < DBC = < DBS

Next we will show that < DBS = 45.

In fact, triangle BSD is an isosceles right triangle with BS = DS and < BSD = 90. Its proof is as follows:

As triangle BFS and triangle DES are congruent,
BS = DS, and
< BSF = < SDE

Because QS and DE are parallel,
< QSD = < SDE

Thus
< BSD = < BSQ + < QSD
= < BSQ + < BSF
= < FSQ
= 90