In triangle ABC, DE is parallel to BC and AD:DB=5:4 diagonals DC and BE intersect at F Find - Maths - Triangles

Question

In triangle ABC, DE is parallel to BC and AD:DB=5:4 diagonals DC
and BE intersect at F Find Area(triangle DEF) / Area(triangle
CFB)

Ankush Jain · Accepted Answer

Given, DE || BC and AD : DB = 5 : 4 Now, âˆ ADE = âˆ ABC and âˆ AED = âˆ ACB [Corresponding angles] In â–³ ADE and â–³ ABC, we have âˆ ADE = âˆ ABC, âˆ AED = âˆ ACB and âˆ A = âˆ A [Common] So, â–³ ADE ~ â–³ ABC [By AAA similarity] â‡’ $\frac{A D}{A B} = \frac{D E}{B C}$ Given, $\frac{A D}{D B} = \frac{5}{4}$ â‡’ $\frac{D B}{A D} = \frac{4}{5}$ Now, adding 1 on both sides,we get $\frac{D B}{A D} + 1 = \frac{4}{5} + 1$ â‡’ $\frac{A B}{A D} = \frac{9}{5}$ or $\frac{A D}{A B} = \frac{9}{5} = \frac{D E}{B C}$ Again, In â–³ DEF and â–³ BCF, we have âˆ EDF = âˆ BCF [Alternate interior angles] âˆ DFE = âˆ BFC [Vertically opposite angles] and âˆ DEF = âˆ FBC [Alternate interior angles] So, â–³ DEF ~ â–³ BCF [By AAA similarity] â‡’ $\frac{A r e a (Δ D E F)}{A r e a (Δ B C F)} = {(\frac{D E}{B C})}^{2}$ â‡’ $\frac{A r e a (Δ D E F)}{A r e a (Δ B C F)} = {(\frac{5}{9})}^{2} = \frac{25}{81}$

In triangle ABC, DE is parallel to BC and AD:DB=5:4. diagonals DC and BE intersect at F. Find Area(triangle DEF) / Area(triangle CFB).