∴∠ADP=∠CDQ.
At △ADP and △CDQ,
Democratic Action Party =∠DCQ=90 degrees
AD=CD
∠ADP=∠CDQ
∴△ADP≌△CDQ(ASA),
∴DP=DQ.
(2) conjecture: PE = QE.
Proof: from (1), we can know that DP = DQ.
At △DEP and △ dirk,
DP=DQ
Partial differential equation =∠QDE=45
DE=DE
∴△DEP≌△DEQ(SAS),
∴PE=QE.
(3) Solution: ∫AB:AP = 3:4, AB=6,
∴AP=8,BP=2.
In the same way as (1), it can be proved that △ ADP △ CDQ,
∴CQ=AP=8.
In the same way as (2), it can be proved that △ DEP △ DEQ,
∴PE=QE.
Let QE=PE=x, then be = BC+CQ-QE =14-X.
In Rt△BPE, from Pythagorean theorem, BP2+BE2=PE2,
Namely: 22+( 14-x)2=x2,
Solution: x=
50
seven
, that is, QE=
50
seven
.
∴S△DEQ=
1
2
QE? CD=
1
2
×
50
seven
×6=
150
seven
.
∫△DEP?△ Dirk,
∴S△DEP=S△DEQ=
150
seven
.
Please adopt it. Thank you.