please answer......... urgent
Attachments:
Answers
Answered by
0
since a triangle is congruent to another triangle..
they are parallelogram..
I know my answer is wrong....
bt please thank me....plzz.....
they are parallelogram..
I know my answer is wrong....
bt please thank me....plzz.....
kinkyMkye:
not congruent but similar
Any triangle is defined by six measures (three sides, three angles)
triangles are similar if
AAA (angle angle angle)
All three pairs of corresponding angles are the same.
aaa in same proportion (side side side)
All three pairs of corresponding sides are in the same proportion
aAa (side angle side)
Two pairs of sides in the same proportion and the included angle equal.
All three pairs of corresponding angles are the same.
aaa in same proportion (side side side)
All three pairs of corresponding sides are in the same proportion
aAa (side angle side)
Two pairs of sides in the same proportion and the included angle equal.
Answered by
0
triangles are similar if aaa in same proportion (side side side)
All three pairs of corresponding sides are in the same proportion
converse of side splitter theorem is
If a line intersects two sides of a triangle and divides the sides proportionally, the line is parallel to the third side of the triangle.
∵ AD/AB=AF/AC
∴ ΔADF~ΔABC and applying converse of side splitter theorem
∴ DF║BC
∵ OP/OB=OQ/OC
∴ ΔOPQ~ΔOBC and applying converse of side splitter theorem
∴ PQ║BC
so DF║PQ................2 sides are parallel
∵ BD/BA=BP/BO
∴ ΔDBP~ΔABO and applying converse of side splitter theorem
∴ DP║AO
∵ CF/CA=CQ/CO
∴ ΔCFQ~ΔACO and applying converse of side splitter theorem
∴ FQ║AO
so DP║FQ................2 sides are parallel
Hence proved
All three pairs of corresponding sides are in the same proportion
converse of side splitter theorem is
If a line intersects two sides of a triangle and divides the sides proportionally, the line is parallel to the third side of the triangle.
∵ AD/AB=AF/AC
∴ ΔADF~ΔABC and applying converse of side splitter theorem
∴ DF║BC
∵ OP/OB=OQ/OC
∴ ΔOPQ~ΔOBC and applying converse of side splitter theorem
∴ PQ║BC
so DF║PQ................2 sides are parallel
∵ BD/BA=BP/BO
∴ ΔDBP~ΔABO and applying converse of side splitter theorem
∴ DP║AO
∵ CF/CA=CQ/CO
∴ ΔCFQ~ΔACO and applying converse of side splitter theorem
∴ FQ║AO
so DP║FQ................2 sides are parallel
Hence proved
Attachments:
Similar questions