Albert1 Messages 1,221 Reaction score 0 Thread starter Jul 6, 2015 #1 find AB:BC:CD Attachments ratio of three segments.jpg 13.4 KB · Views: 107
Albert1 Messages 1,221 Reaction score 0 Jul 10, 2015 #2 Albert said: find AB:BC:CD hint : Spoiler use law of sine,and find the ratio of area
Albert1 Messages 1,221 Reaction score 0 Jul 10, 2015 #3 Albert said: hint : Spoiler use law of sine,and find the ratio of area more hint : Spoiler (1) find area of triangle ABP:CDP (2) find area of triangle ACP:BDP
Albert said: hint : Spoiler use law of sine,and find the ratio of area more hint : Spoiler (1) find area of triangle ABP:CDP (2) find area of triangle ACP:BDP
Albert1 Messages 1,221 Reaction score 0 Jul 11, 2015 #4 Albert said: find AB:BC:CD solution of others Spoiler $\dfrac {\triangle ABP}{\triangle CDP}=\dfrac {12\times 9}{10\times 16}=\dfrac {27}{40}=\dfrac {AB}{CD}$ let $AB=27t, CD=40t,BC=k$ $\dfrac {\triangle APC}{\triangle BPD}=\dfrac {12\times 10}{9\times 16}=\dfrac {27t+k}{40t+k}=\dfrac {5}{6}$ we have $k=38t$ and $AB:BC:CD=27:38:40$
Albert said: find AB:BC:CD solution of others Spoiler $\dfrac {\triangle ABP}{\triangle CDP}=\dfrac {12\times 9}{10\times 16}=\dfrac {27}{40}=\dfrac {AB}{CD}$ let $AB=27t, CD=40t,BC=k$ $\dfrac {\triangle APC}{\triangle BPD}=\dfrac {12\times 10}{9\times 16}=\dfrac {27t+k}{40t+k}=\dfrac {5}{6}$ we have $k=38t$ and $AB:BC:CD=27:38:40$