The median AD of a triangle ABC is produced and a perpendicular CF is dropped on it. BE is perpendicular to AD. If BC=34cm and DF=8cm, what is the length (in cm) of BE?
- A17
- B9
- C15
- D19
Solution & Step-by-step Explanation
In △ABC, AD is a median, which means D is the midpoint of BC.
BD=CD=
2
BC
=
2
34
=17cm
Consider triangles △BED and △CFD:
∠BED=∠CFD=90
∘
(given that BE⊥AD and CF⊥AD)
∠BDE=∠CDF (vertically opposite angles)
BD=CD=17cm (since D is the midpoint)
By AAS (Angle-Angle-Side) congruence criterion:
△BED≅△CFD
By CPCT (Corresponding Parts of Congruent Triangles):
BE=CF
ED=DF=8cm
Now apply Pythagoras' theorem in the right-angled triangle △CFD (where ∠CFD=90
∘
):
CD
2
=CF
2
+DF
2
17
2
=CF
2
+8
2
289=CF
2
+64
CF
2
=289−64=225
CF=
225
=15cm
Since BE=CF, the length of BE is 15cm.
BD=CD=
2
BC
=
2
34
=17cm
Consider triangles △BED and △CFD:
∠BED=∠CFD=90
∘
(given that BE⊥AD and CF⊥AD)
∠BDE=∠CDF (vertically opposite angles)
BD=CD=17cm (since D is the midpoint)
By AAS (Angle-Angle-Side) congruence criterion:
△BED≅△CFD
By CPCT (Corresponding Parts of Congruent Triangles):
BE=CF
ED=DF=8cm
Now apply Pythagoras' theorem in the right-angled triangle △CFD (where ∠CFD=90
∘
):
CD
2
=CF
2
+DF
2
17
2
=CF
2
+8
2
289=CF
2
+64
CF
2
=289−64=225
CF=
225
=15cm
Since BE=CF, the length of BE is 15cm.