Geometry

Triangles

Hard

In a triangle ABC, $\angle BCA = 50^\circ$ . 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 $\angle FDE$ , in degrees, is equal to

100

✅ Correct Option: 4

Triangle ADE is isosceles because AD = DE (given). In any isosceles triangle, the base angles are equal.

So $\angle DAE = \angle DEA = x$ (let's call this angle x)

The third angle: $\angle ADE = 180° - 2x$

Triangle BDF is isosceles because BD = DF (given). Similarly, $\angle DBF = \angle DFB = y$ (let's call this angle y)

The third angle: $\angle BDF = 180° - 2y$

We observe that:

$\angle CAB = \angle DAE = x$ (same angle!)

$\angle ABC = \angle DBF = y$ (same angle!)

$\angle BCA = 50°$ (given)

Since angles in triangle ABC sum to 180°:

$\angle CAB + \angle ABC + \angle BCA = 180°$

$x + y + 50° = 180°$

Therefore: $x + y = 130°$

At point D, we have three angles meeting:

$\angle ADE$ (from triangle ADE)

$\angle BDF$ (from triangle BDF)

$\angle FDE$ (what we want to find)

These three angles form a straight line along the side of triangle ABC, so they sum to 180°.

$\angle ADE + \angle BDF + \angle FDE = 180°$

$(180° - 2x) + (180° - 2y) + \angle FDE = 180°$

$360° - 2x - 2y + \angle FDE = 180°$

$\angle FDE = 180° - 360° + 2x + 2y$

$\angle FDE = 2x + 2y - 180°$

$\angle FDE = 2(x + y) - 180°$

Since $x + y = 130°$ :

$\angle FDE = 2(130°) - 180° = 260° - 180° = 80°$

$\angle FDE = 80°$

This problem beautifully combines isosceles triangle properties with angle relationships. The key was recognizing that the angles x and y in our isosceles triangles are actually the same as angles A and B in the main triangle ABC.

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In a triangle ABC, ∠BCA=50∘\angle BCA = 50^\circ∠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\angle FDE∠FDE, in degrees, is equal to

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CAT Past Year Questions (Topic-Wise)

In a triangle ABC, $\angle BCA = 50^\circ$ . 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 $\angle FDE$ , in degrees, is equal to