in ABC, EF is parallel to AC if EF is equal to 2 and AC is equal to 5 find BE is to EA
Answers
Answered by
0
ANSWER
Draw a perpendicular from C onto AB to meet AB at G and EF at H
Area △ABC=1/2×AB×CG
=ar(EFBA)+ar(CEF)
=2×△CEF=EF×CH (because ar(EFBA)=ar(CEF) and ar(CEF)=1/2×CH×EF)
AB∗CG=2×EF×CH.....(i)
△ABC and △CEF are similar triangles. So, the ratios of their corresponding sides and altitudes are equal.
EF/AB=CE/CA=CF/CB=CH/CG
CH=EF∗CG/AB
So (1)=>AB∗CG=2 EF ^2
∗CG/AB
AB^ 2
=2 EF^ 2
AB√2EF
AB/EF=CA/CE=√2
The ratio of the sides in the two triangles is √2
CA=√2CE
CA/EA=CA/(CA−CE)=√2CE/(√2CE−CE)
CA/EA=√2/(√2−1)=√2(√2+1)
Similar questions