since AP=BQ=CR
ABCD is a square
so PB=QC
angle PBQ=angle QCR
so triangle PBQ is congruent to triangle QCR
so PQ =QR
so PQR is an isos. triangle 1.
a. Triangle QXY is similar to triangle QPR;
=> angle QXY = angle QPR, thus XY // PR
also, triangle AXY is similar to triangle ABC;
=> angle AXY = angle ABC, thus XY // BC
PR // XY // BC and hence the result.
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