Mathematics Textbook Grade 8 Volume II Answer p68 Question 13

1, PQ∑CD, then the quadrilateral CDPQ is a parallelogram,

PQDC is a parallelogram,

∴PD=QC，

PD=AD-AP=24-t，

QC=3t，

∴24-t=3t, and the solution is: t = 6；;

2.① Same as above: PQDC is parallelogram.

∴PD=QC,PD=AD-AP=24-t，

QC=3t, ∴24-t=3t, and the solution is: t=6.

② Quadrilateral CDPQ is isosceles trapezoid:

The PH⊥BC in h exceeds p, and the DE⊥BC in e exceeds d,

PQCD is an isosceles trapezoid,

∴PQ=DC,∠PQH=∠DC，

∵PH⊥BC,DE⊥BC,∴∠PHQ=∠DEC=90

* in △PHQ and △1February.

∠PQH=∠DCE, ∠PHQ =∠ dirk, PQ=DC,

∴△PHQ≌△DEC(AAS)，

∴HQ=CE,CE=BC-AD=26-24=2，

HQ=QC-PD-CE=3t-(24-t)-2=4t-26，

4t-26=2，t=7。