Question

in triangle abc d e f are midpoints of sides AC ab and BC respectively BG is an altitude of triangle ABC if angle EDF is equal to 56 degree then find the measure of angle EGF in degrees​

Answer 1

Answer:

Step-by-step explanation:

(1)  Since E and F are the midpoints of AC and AB.

BC||FE & FE= ½ BC= BD

BD || FE & BD= FE         ( By mid point theorem)

Similarly,

BF||DE & BF= DE

   Hence, BDEF is a parallelogram                

(2)  Similarly, we can prove that FDCE & AFDE are also parallelogram

Since,

BDEF is a parallelogram so its diagonal FD divides its into two Triangles of equal areas.

∴ ar(ΔBDF) = ar(ΔDEF)      ......... (1)

In Parallelogram AFDE

ar(ΔAFE) = ar(ΔDEF) (EF is a diagonal)      ......... (2)

In Parallelogram FDCE

ar(ΔCDE) = ar(ΔDEF) (DE is a diagonal)      ......... (3)

From (1), (2) and (3)

ar(ΔBDF) = ar(ΔAFE) = ar(ΔCDE) = ar(ΔDEF)      .....(4)

ar(ΔBDF) + ar(ΔAFE) + ar(ΔCDE) + ar(ΔDEF) = ar(ΔABC)

4 ar(ΔDEF) = ar(ΔABC)         (From eq 4)

   ar(∆DEF) = 1/4 ar(∆ABC)      ........(5)

(3)  Area (parallelogram BDEF) = ar(ΔDEF) + ar(ΔBDF)ar(parallelogram BDEF) = ar(ΔDEF) + ar(ΔDEF)

ar(parallelogram BDEF) = 2× ar(ΔDEF)         (From eq 4)

ar(parallelogram BDEF) = 2× 1/4  

ar(ΔABC)          (From eq 5)

   ar(parallelogram BDEF) = 1/2 ar(ΔABC)

Answer 2

The factor theorem is used in mathematics to connect the factor and zeros of a polynomial. The factor theorem is frequently used to factor a polynomial and find the roots of a polynomial equation. It is a subset of the polynomial remainder theorem.

How to solve the theorem?

Consider the unknown. To find the value of X, use the Pythagorean theorem. Your answer should be rounded to the nearest hundredth. Remember our steps for applying this theorem. This problem is similar to Example 2 in that we are solving for one of the legs. Determine the legs and hypotenuse of a right triangle.

(1) Because the midpoints of AC and AB are E and F

BC||FE & FE= 12 BC=

BD = FE and BD = FE (According to the midpoint theorem)

Similarly

BF||DE and BF=DE

As a result, BDEF is a parallelogram.

2) Similarly, we can demonstrate that FDCE and AFDE are parallelograms.

Since,

Because BDEF is a parallelogram, its diagonal FD divides it into two equal-area triangles.

ar(BDF) = ar(DEF)........... (1)

AFDE Parallelogram

ar(AFE) = ar(DEF) (EF is a diagonal).......... (2)

FDC Parallelogram

To learn more about the theorem.

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