Question

A triangle ABC has sides AB=50 and AC=10. D is mid-point of AB and E is mid-point of AC. Angle bisector AG from vertex A meets side BC at G and divides it in 5:1 ratio. So BG=5 times GC. AG cuts ED at F. Find the ratio of the areas of the trapeziods FDBG to FGCE.

Answer 1

Ans 5:1   (Refer the solution below)

Consider , l(AB)=50 , l (AC) = 10 .............. (1)

BG : GC = 5 : 1 ................(given)

Assume BG = 5x ...... (2)            GC = x .............. (3)

Now Pt D is the midpoint of ............ (4)

Pt E is the midpoint of ............ (5)

........... (triangle Midpoint theorem)............(6)

i.e.

We can easily prove .............. (BY A-A test of similarity i.e. common angle and corresponding angle test)

...................... (C.S.S.T)

.............(9)

we can prove ................. (10)

Let us construct

........(By definition of parallel lines) .............. (11)

From 6 and 10 quadrilateral DHIE is a parallelogram

............... (Opp sides)................ (12)

Now Area of trapezoid FDBG = ..........(13)

and Area of trapezoid FGCE = .... (14)

...... (Due to Stmnts 12, 13 and 14)

Ans: Ratio of trapezoids FDBG to FGCE = 5: 1