Thursday

November 27, 2014

November 27, 2014

Posted by **Anonymous** on Saturday, January 15, 2011 at 2:46am.

- geometry -
**Reiny**, Saturday, January 15, 2011 at 9:11amLets concentrate on the quadrilateral ADEG.

let the intersection of AE and DB be K

GE || AD (given)

So triangles ADE and ADG have the same base AD and the same height , since parallel lines are equal distances apart.

So triangle ADE = triangle ADG

but ADK is common to both, so subtracting ...

triangle AKG = triangle DEK

similarly by letting FD and BC intersect at L we can show that triangle FLB = triangle CDL

pentagon ABCDE

= triangle BLD +**triangle LCD**+ triangle ABD +**triangle DEK**+ triangle ADK

= triangle BLD +**triangle FLB**+ triangle ABD +**triangle GAK**+ triangle ADk

= triangle FDG

**Answer this Question**

**Related Questions**

Math - Two circles C1 and C2 touch externally at A. The tangent to C1 drawn at a...

maths - ABCDE is a Pentagon with BE parallel to CD and BC parallel to DE. BC is ...

maths - The convex pentagon ABCDE has a right angle at B,AB=BC andCD=DE=EA=1cm,...

MATHS - two circles c1 and c2 meet at the points A and B. CD is a common tangent...

maths-circles - ABCD is a rectangle the line through C perpendicular to AC meets...

geometry - Which geometric figure could not be drawn using both perpendicular ...

geometry - Find the area of a pentagon ABCDE with verticles A(0,4),B(3,2),C(3,-1...

geometry - Find the area of a pentagon ABCDE with verticles A(0,4),B(3,2),C(3,-1...

Geometry - Prove that the tangents to a circle at the endpoints of a diameter ...

analytical geometry - the axes are rectangular and a point P moves on the fixed ...