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 .
Hello,
For that, I will use a rough method, using more algebra than geometry.
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Because I don't remember the formula for , I'll show it.
~~~~~~~~~~~~~~
For that I will need to find the formula (which I don't remember either) of tan(c+d) :
Let's multiply by
This gives :
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Therefore, we have :
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Now let's go back to the problem.
So we have
(formula above).
Therefore,
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