TheCATExam
Sign in

CAT 2021 Slot 3 — QA Question 2

Basics of TrianglesEasy

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

Answer & solution

  • A

    96

  • B

    100

  • 80

  • D

    72

Solution

Easy

DD sits on ABAB with two isosceles triangles hanging off it: △ADE\triangle ADE (since AD=DEAD=DE) and △BDF\triangle BDF (since BD=DFBD=DF). The base angles of those triangles equal ∠A\angle A and ∠B\angle B. Since A,D,BA,D,B are collinear, the three angles at DD add to a straight angle, which pins down ∠FDE\angle FDE.

A B C D E F ∠FDE
1

Name the base angles. Let ∠A=x\angle A=x and ∠B=y\angle B=y. The angle sum of △ABC\triangle ABC with ∠C=50∘\angle C=50^\circ gives a relation between xx and yy.

x+y+50∘=180∘(angle sum of △ABC)⇒ x+y=130∘\begin{aligned} &x+y+50^\circ=180^\circ \quad\text{(angle sum of }\triangle ABC\text{)}\\ &\Rightarrow\ x+y=130^\circ \end{aligned}
2

Use the isosceles triangle at AA. Since AD=DEAD=DE, triangle ADEADE is isosceles with base angles equal to ∠A\angle A.

∠DEA=∠DAE=x(AD=DE)⇒ ∠ADE=180∘−2x(angle sum of △ADE)\begin{aligned} &\angle DEA=\angle DAE=x \quad\text{(}AD=DE\text{)}\\ &\Rightarrow\ \angle ADE=180^\circ-2x \quad\text{(angle sum of }\triangle ADE\text{)} \end{aligned}
3

Use the isosceles triangle at BB. Since BD=DFBD=DF, triangle BDFBDF is isosceles with base angles equal to ∠B\angle B.

∠DFB=∠DBF=y(BD=DF)⇒ ∠BDF=180∘−2y(angle sum of △BDF)\begin{aligned} &\angle DFB=\angle DBF=y \quad\text{(}BD=DF\text{)}\\ &\Rightarrow\ \angle BDF=180^\circ-2y \quad\text{(angle sum of }\triangle BDF\text{)} \end{aligned}
4

Straight angle at DD. Points A,D,BA,D,B are collinear, so the three angles ∠ADE\angle ADE, ∠FDE\angle FDE, ∠BDF\angle BDF sit along a straight line.

∠ADE+∠FDE+∠BDF=180∘⇒ (180∘−2x)+∠FDE+(180∘−2y)=180∘(from steps 2, 3)⇒ ∠FDE=2x+2y−180∘⇒ ∠FDE=2(130∘)−180∘=80∘(from step 1)\begin{aligned} &\angle ADE+\angle FDE+\angle BDF=180^\circ\\ &\Rightarrow\ (180^\circ-2x)+\angle FDE+(180^\circ-2y)=180^\circ \quad\text{(from steps 2, 3)}\\ &\Rightarrow\ \angle FDE=2x+2y-180^\circ\\ &\Rightarrow\ \angle FDE=2(130^\circ)-180^\circ=80^\circ \quad\text{(from step 1)} \end{aligned}
∠FDE=80∘\angle FDE=80^\circ