∴∠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.