Find The Interior Angle

Triangle ABC is an isosceles triangle. <A = 80. Therefore, <B =<C =50 degrees.

Since triangle CDE is also isosceles, and <C =50, <CDE = <CED = 65 degrees.

Also, triangle DFB is isosceles, <B =50. Therefore, <FDB = <DFB = 65 degrees.

<CDE, <EDF, < FDB are angles on a straight line with <CDE = <FDB =65 degrees. Therefore, <EDF = 180 - 65 - 65 = 50 degrees.

If the sides AB and BC are congruent, it is an isosceles triangle. As the angle at the vertice A is 80 degrees, then the angles at the vertices B and C is 50 degrees each. Putting the D, E and F points on their respective sides, we realize that formed the CDE, DEF and BDF triangles that are also isosceles. As the sum of the interior angles of all triangles is equal to 180 degrees, we know that the angles at the vertices D, E and F is 65 degrees. The EFD angle is equal to 180 - (65 + 65) degrees. Therefore we conclude that this angle is 50 degrees. Sorry about the grammatical errors... I really tried to write the best I could. Thank you! ;D

ATQ, -angB = angC = 1/2(180-80) = 50 (Since AB=AC & applying angle sum property)

-In triangleBFD, angBFD = angBDF = 65 (-------------)

-In triangleCED, angCED = angCDE = 65 (-------------)

-Also, angBDF + angEDF + angCDE = 180 (Straight angle)

-So, angEDF = 180- ( angCDE+ angBDF) = 180 - (65 + 65)

angEDF = 50

A = 80 and it is given that ac = ab

In triangle ABC,

Let C and B be x as they are equal(angle opposite to equal sides are equal)

A + x + x = 180

80 + 2x = 180

x = 50

In triangle CED ,

Let angle CED and angle EDC be y ,

C + y + y = 180

50 + 2y = 180

y = 65

thus, angle EDC = 65

Similarly angle FDB = 65

angle CDB = 180

CDE + EDF + BDF = 180

65 + EDF + 65 = 180

EDF = 180 - 130

Hence angle EDF = 50

at first i have drawn two perpendicular from the point D to AB and AC ,respectively DP and DQ . Now angleDPB=90 angleDBP=50(180-80/2) so anglePDB=40 similarly angleQDC=40 so angleQDP =100 angleDFB=angleBDF=65 so angleFDP=25(90-65) similarly angleQDE=25 so angleEDF=50(100-50)

Triangle ABC is an isosceles triangle

A=80 B=C=50

Triangles FBD and EDC are congruent and isosceles triangles

E=D=F=65

Straight Angle is equal to 180

thus angle EDF = 180-130= 50

In triangle ABC,A=80 degrees.CE=CD &BF=BD,therefore,[angles opp equal sides are equal]angles CDE & CED[x] are equal.similarly,angles BDF=BFD[y].Angle A=180-2x & angleB=180-2y.By the angle sum property,80+180-2x+180-2y+180.So 2x+2y=260.So x+y=130.Therefore,x+y+EDF=180[linear pair].So angle EDF=50 degrees

AB=AC, <CAB=80 Degree so,<ACB=80

Triangle CED=<ECD+<CDE+CED

<ECD=80 Degree

so <CDE+<CED=100

and also <CDE=<CED

so <CDE=<CED =50 Degree

At the same time

CDB=180 Degree

CDB includes <CDE+<EDF<FDB

So EDF=50 Degree

triangle ABC is an isosceles triangle if angle A=80 then angle B&C=100; 50 each. triangle BDF is also an isosceles triangle since BF=BD ; angle B=50 then angle F and Angle BDF is equal=130/2 = 65 each; traingle ECD is an isosceles triangle also the same as triangle BFD angle E and EDC is 65 each; CDB=180 less the 65 for angle BDF and 65 angle CDE EFF=180-65-65=50

It is very easy one . just try to calculate all angles in all triangles using congruent angles and sides :D

because sides AB=AC, then angle ACB=angle ABC=X(LET), GIVEN that angle CAB=80 degree, so in triangle ABC, 2X+80=180,(because in a triangle sum of three angles=180) so,X=50, GIVEN IN triangle CED, CE=CD, SO,angle CED= angle CDE=(180-50)/2=65 degree. given IN triangle BDF, BF=BD. SO, angle BDF= angle BFD=(180-50)/2=65 degree. now , angle BDF+ angle FDE+angle CDE=180 or, 65+angle FDE+65=180, SO,angle FDE=50 degree.

In Triangle ABC <A=80 &AB=AC Therefore <B=<C By Angle Sum Property Of a Triangle <A+ <B+ <C=180 80+<b+<c=180 2<B=100 <B=100=<C
Similarly, <cde =<e=65 &<fdb =<dfb=65 <CDE+<BDF+<EDF=180 <EDF=180-65-65 =180-130=50 Therefore <EDF=50

As we know that "if two sides of a triangle r equal so the angles opposite to the sides will also be equal". so as in triangleABC AB=AC so AngleC=AngleB since AngleA=80 So, AngleC/B= 180-80\2 = 100\2 =50 Hence AngleB=AngleC=50

now in triangle CED, AngleC=50degree As CE=CD so AngleE=AngleD= 180-50\2=65

similarly in TriangleDFB, AngleB-50degree As DF=BD so, AngleD=AngleF=180-50\2=65

As AngleCDB=180 So AngleEDF= 180-(angleCDE+angleBDF) By putting the values we get, AngleEDF=180-(65+65) AngleEDF=180-130=50

Hence the value of AngleEDF is 50degree.

With 80 degree angle other two anles are 50 degree each( being isoscles triangle) similarly,for other two triangles,angles adjacent to angle EDF are 65 degree,Sum of all angles is 180 degree therefore angle EDF is 180 minus(65 +65)+50 degree

since AB=AC,therefore Triangle ABC is isosceles Triangle. Given that Angle A=80,hence Angle B= Angle C= 50 . Similarly Triangles CED & BFD are also isosceles Triangle .Therefore angle CED= angle CDE= angle BFD= angle BDF=65 degree.Hence angle EDF=50 degree

here given that ab=ac <abc+<acb+<a=180 <acb = <abc = 50 similarly, <ced = <cde = 65 (given ce=cd) <bfd =<bdf =65 (given bf=bd) <cdb=180 =<cde+edf+bdf <cde+<bdf = 130 therefore <EDF =50

A+B+C=180,B=C B+C=180-80=100 B=C=50 E+D+C=180,E=D E=D=65 D+B+F=180,D=F D=F=65 EDC+EDF+FDB=180 EDF+130=180 EDF=50

Since ABC is isosceles then C = B = 50 . since CDE and FBD are isosceles then E = D = F = 65, Clearly, EDC, FDB and EDF sum up to 180 so EDF is 50 degrees

Since △ABC is isosceles, ∠ACB=∠ABC=\frac {180∘-80∘}{2}= 50∘. △BFD and △ECD are both isosceles triangles with equal top angles, and so ∠CED=∠CDE=∠FBD=∠FDB= \frac {180∘-50∘}{2}= 65∘. ∠EDF= 180∘-(∠FDB+∠EDC)= 50∘.

That's it.

In Δ ABC AB=AC , ∠A=50° ⇒ ∠C=∠B=(180-80)/2=50°

In Δ CED CE=CD , ∠C=50° ⇒ ∠CED=∠CDE=(180-50)/2=65°

In Δ FDB BF=BD , ∠DBF=50° ⇒ ∠BDF=∠BFD=(180-50)/2=65°

∠CDB=180° ( straight angle ) ⇒∠EDF= 180-(65+65)=180-130= 50°

For this explanation, I used uppercase for sides and lowercase for angles to prevent confusion.

if AB = AC and a = 80 degrees, then b = c

$a + b + c = 180$

$a + 2b = 180$

$2b = 180 - 80$

$2b = 100$

$b = 50$

Since CE = CD, we could say that d = e for CDE. Thus,

$b = 180 - 2d$

$2d = 180 - b$

$2d = 180 - 50$

$2d = 130$

$d = 65$

the same could be said for d and f for BDF.

From there,

$cde + edf + fdb = 180$

$edf = 180 - 65 - 65$

$edf = 180 - 130$

$edf = \boxed{50}$

Since A=80, then B=50 and A= 50 .Since Triangle CDE is isosceles that makes angle D=65 and angle E = 65 and SINCE Triangle BDF is isosceles, that makes angle F = 65 angle D= 65. since angle D from both triangles sum to 130, therefore EDF is 50.

Ang A = 80 since AB = AC then Ang B = Ang C So Ang C = Ang C = 50 Since CE = CD So Ang CED + Ang CDE = 130 (since Ang C = 50) Ang CED = Ang CDE = 65 (Vice Versa) similarly Ang BFD = Ang BDF = 65 Ang BDF + Ang CDE + Ang EDF = 180 Hence Ang EDF = 50

Triangle ABC is an isosceles triangle, AB = AC so angle B and c are the same angle that is (180 - (point A)): 2 = 50. Both angles B and C is 50. Then look at triangle CDE is also an isosceles triangle, so CD = CE angle D and E have the same value that is (180 - (point C)): 2 = 65. Similarly, the triangle BDF. Because the angle of the triangle CDE D has a value equal to the angle D in the triangle BDF, look straight line BC is worth 180 degrees. Then the angle EDF is (180 - (2 x 65)) = 50.

Since AB and AC are equal, therefore angle ABC and angle ACB will be equal to each other. Using the angle sum property of a triangle each of the 2 angles can be found out to be 50 degrees each. Proceeding in a similar way in triangles CED and BDF we can get the equal angles of the triangle as 65 degrees each. again the sum of the angles CDE, EDF and BDF will be 180 degrees. Therefore 65+EDF+65 =180... this gives us EDF =50

triangle abc is isosceles as AB=AC. so angle B=angle C=180-80/2 . triangle CED and DFB are also isosceles.solving we get angle EDF as 50 degrees