Triangle PDQ is similar to triangle PBA.
therefore
PD:PB==PQ:PA
A triangular PDA is similar to a triangular PBR.
therefore
PD:PB=PA:PR
Multiply the two formulas to get it.
PQ:PR=PD^2:PB^2
2. As shown in the figure, it is known that AD is the center line of △ABC, and a straight line passes through AB in E, AD in G and AC in F. ..
Verify AB/AE+AC/AF=2AD/AG.
Point D is considered as MN parallel EF, intersecting with AB at M point, intersecting with AC extension line at N point, CK parallel with AB, and intersecting with MN at K point.
It is easy to prove the triangles AMD∽AEG, ADN∽AGF, CKN∽AEF, MBD≌KCD.
Therefore AM/AE=AD/AG, AN/AF=AD/AG, CK/AE=NC/AF, BM=CK.
Therefore (AB/AE)+(AC/AF)
=[(AM+BM)/AE]+[(AN-NC)/AF]
=(AM/AE)+(AN/AF)+[(BM/AE)-(NC/AF)]
=(AM/AE)+(AN/AF)+[(CK/AE)-(NC/AF)]
=(AM/AE)+(AN/AF)
= (advertising/advertising)+(advertising/advertising)
=2AD/AG