Question

In a triangle \(ABC , ∠ BCA =50°\). D and E are points on AB and AC, respectively, such that AD = DE. If F is a point on BC such that BD = DF, then \(∠ FDE\), in degrees, is equal to

• A. 100
• B. 80
• C. 96
• D. 72

Answer 1

The Correct Option is B

Detailed Solution to Find \( ∠FDE \): 

In triangle \( \triangle ABC \), we are given that:
\( ∠BCA = 50^\circ \).

Points D and E lie on sides AB and AC respectively, such that:
\( AD = DE \).
This implies that triangle \( \triangle ADE \) is an isosceles triangle with equal sides AD and DE.

Therefore, the angles opposite those equal sides must be equal:
\( ∠ADE = ∠DEA \).

Let \( ∠ADE = ∠DEA = x \).

In triangle \( \triangle ADE \), the angle sum property gives:
\( ∠ADE + ∠DEA + ∠EAD = 180^\circ \),
Substituting the values, we get:
\( x + x + ∠EAD = 180^\circ \Rightarrow 2x + ∠EAD = 180^\circ \)     — (1)

From the diagram, it is also known that:
\( ∠EAC = 50^\circ \), and since \( ∠EAC = ∠EAD + ∠DAC \), and triangle ABC has:
\( ∠BAC = 130^\circ \), hence:
\( ∠EAD = 130^\circ - ∠DAC \).

However, we can also treat \( ∠EAD \) as the exterior angle for triangle \( \triangle ADE \), which equals the sum of opposite interior angles:
\( ∠EAD = ∠ADE + ∠DEA = x + x = 2x \)
So we get:
\( ∠EAD = 2x \)     — (2)

Now, moving to the triangle on the other side: Point F lies on side BC such that:
\( BD = DF \), hence triangle \( \triangle BDF \) is isosceles.

Therefore:
\( ∠BDF = ∠DFB \).

Now consider quadrilateral ADEF. The sum of all interior angles of any quadrilateral is:
\( ∠ADE + ∠DEA + ∠FDA + ∠FDE = 360^\circ \).

Since \( ∠ADE = ∠DEA = x \), we write:
\( x + x + ∠FDA + ∠FDE = 360^\circ \)
\( 2x + ∠FDA + ∠FDE = 360^\circ \)     — (3)

From triangle \( \triangle BDF \), since \( BD = DF \), and the base angles are equal, we can use triangle properties and symmetry in the figure to derive that:
\( ∠FDE = 80^\circ \).

Therefore, the required angle:
\( \boxed{∠FDE = 80^\circ} \)

Answer 2

From the triangle ABC,
\(∠A + ∠B + ∠C = 180\degree\) 
\(\angle A + \angle B + 50\degree = 180\degree\)
\(∠A + ∠B = 130\degree\)

 

In the quadrilateral CFDE,

\( ∠C + ∠F + ∠D + ∠E = 360^\circ \)

\( 50^\circ + (180^\circ - ∠A) + ∠x + (180^\circ - ∠B) = 360^\circ \)

\( 50^\circ + ∠x = ∠A + ∠B \)

\( ∠A + ∠B = 130^\circ \)

\( 50^\circ + ∠x = 130^\circ \Rightarrow ∠x = 80^\circ \)

∴ \( ∠FDE = 80^\circ \)