In triangle abc which is isosceles a b equals a c and the measure of angle bac is equal to 20 degrees on side a b drop point d such that a d is equal to bc find the measure of angle dca

  ∠DCA = 10°

Step-by-step explanation:

i know of no simple way to show the angle is 10° and that ΔAFC is isosceles (see the attached figure). Using the Law of Sines with respect to ΔADC, you can show that ∠DCA = 10°.

The measure of side AC is ...

  AC·cos(80°) = EC

The measure of side AD is ...

  AD = 2·EC

  AD = 2·AC·cos(80°) . . . . . . . substituting for EC

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Another relation involving these sides comes from the law of sines.

  sin(∠DCA)/AD = sin(∠ADC)/AC

The sum of angles in a triangle is 180°, so ...

  ∠DCA + ∠ADC + ∠A = 180°

Solving for ∠ADC, we have ...

  ∠ADC = 180° - (∠A + ∠DCA)

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By the symmetry of the sine function, we can rewrite the above law of sines expression as ...

  sin(∠DCA)/(AC·2·cos(80°)) = sin(∠A + ∠DCA)/AC . . . . . . [eq1]

Multiplying by the denominator on the left and expanding the sine of the sum of angles, we have ...

  sin(∠DCA) = 2·cos(80°)·(sin(20°)cos(∠DCA) +cos(20°)sin(∠DCA))

Separating sine and cosine terms, we have ...

  sin(∠DCA)·(1 -cos(20°)·2·cos(80°)) = cos(∠DCA)·sin(20°)·2·cos(80°)

And this gives us an expression for the tangent of ∠DCA:

  tan(∠DCA) = 2·sin(20°)·cos(80°) / (1 -2·cos(20°)·cos(80°))

  arctan(2·sin(20°)·cos(80°) / (1 -2·cos(20°)·cos(80°))) = ∠DCA = 10° (exactly)

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You can simplify the tangent further using various trig identities, but that's more work than we like. You can show the answer is exactly 10° by substituting that value into the rewritten law of sines equation [eq1] above. This gives ...

  sin(10°)/(2·cos(80°)) = sin(30°)

Since cos(80°) = sin(10°), this is a tautology:

  1/2 = sin(30°)

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Possible alternate solution approach

You may be able to do something with the fact that AE is an angle bisector, so DF/DA = CF/CA.