CAT 2021 Slot 3 · Quant · Question 5

Write \angle A=\alpha and \angle B=\beta. Since the angles of the triangle sum to {180}^{∘} and \angle C={50}^{∘}, we have \alpha +\beta ={130}^{∘}.

Triangle ADE is isosceles with AD = DE, so the angles opposite those sides are equal: \angle AED=\angle DAE=\alpha. Hence \angle ADE={180}^{∘}-2\alpha.

Triangle BDF is isosceles with BD = DF, so similarly \angle DFB=\angle DBF=\beta and \angle BDF={180}^{∘}-2\beta.

The three angles at D sit on the straight line AB, so they add to {180}^{∘}:

\angle FDE={180}^{∘}-\left({180}^{∘}-2\alpha \right)-\left({180}^{∘}-2\beta \right)=2\left( \alpha +\beta \right)-{180}^{∘}.

So \angle FDE=2\times{130}^{∘}-{180}^{∘}={80}^{∘}.

Notice that \alpha and \beta never need to be known separately, only their sum.