In figure below, line segment \( \mathrm{DF} \) intersect the side \( \mathrm{AC} \) of a triangle \( \mathrm{ABC} \) at the point \( \mathrm{E} \) such that \( \mathrm{E} \) is the mid-point of \( \mathrm{CA} \) and \( \angle \mathrm{AEF}=\angle \mathrm{AFE} \). Prove that \( \frac{\mathrm{BD}}{\mathrm{CD}}=\frac{\mathrm{BF}}{\mathrm{CE}} \)
[Hint: Take point \( \mathrm{G} \) on \( \mathrm{AB} \) such that \( \mathrm{CG} \| \mathrm{DF} \).]
"
Given:
Line segment \( \mathrm{DF} \) intersect the side \( \mathrm{AC} \) of a triangle \( \mathrm{ABC} \) at the point \( \mathrm{E} \) such that \( \mathrm{E} \) is the mid-point of \( \mathrm{CA} \) and \( \angle \mathrm{AEF}=\angle \mathrm{AFE} \).
To do:
We have to prove that \( \frac{\mathrm{BD}}{\mathrm{CD}}=\frac{\mathrm{BF}}{\mathrm{CE}} \)
Solution:
Take a point $G$ on $A B$ such that $C G \| E F$.
$E$ is the mid-point of $C A$
This implies,
$C E=A E$..........(i)
In $\triangle A C G, C G \| E F$
$E$ is the mid-point of $C A$.
This implies,
$C E=G F$........(ii)
In $\triangle B C G$ and $\triangle B D F, C G \| E F$
By using basic proportionality theorem, we get,
$\frac{B C}{C D}=\frac{B G}{G F}$
$\frac{B C}{C D}=\frac{B F-G F}{G F}$
$\frac{B C}{C D}=\frac{B F}{G F}-1$
$\frac{B C}{C D}+1=\frac{B F}{C E}$ [From (ii)]
$\frac{B C+C D}{C D}=\frac{B F}{C E}$
$\frac{B D}{C D}=\frac{B F}{C E}$
Hence proved.
Related Articles
- In figure below, \( l \| \mathrm{m} \) and line segments \( \mathrm{AB}, \mathrm{CD} \) and \( \mathrm{EF} \) are concurrent at point \( \mathrm{P} \).Prove that \( \frac{\mathrm{AE}}{\mathrm{BF}}=\frac{\mathrm{AC}}{\mathrm{BD}}=\frac{\mathrm{CE}}{\mathrm{FD}} \)."
- \( \mathrm{ABC} \) is a triangle right angled at \( \mathrm{C} \). A line through the mid-point \( \mathrm{M} \) of hypotenuse \( \mathrm{AB} \) and parallel to \( \mathrm{BC} \) intersects \( \mathrm{AC} \) at \( \mathrm{D} \). Show that(i) \( \mathrm{D} \) is the mid-point of \( \mathrm{AC} \)(ii) \( \mathrm{MD} \perp \mathrm{AC} \)(iii) \( \mathrm{CM}=\mathrm{MA}=\frac{1}{2} \mathrm{AB} \)
- In figure below, if \( \mathrm{AB} \| \mathrm{DC} \) and \( \mathrm{AC} \) and \( \mathrm{PQ} \) intersect each other at the point \( \mathrm{O} \), prove that \( \mathrm{OA} \cdot \mathrm{CQ}=\mathrm{OC} \cdot \mathrm{AP} \)."
- In right triangle \( \mathrm{ABC} \), right angled at \( \mathrm{C}, \mathrm{M} \) is the mid-point of hypotenuse \( \mathrm{AB} \). \( \mathrm{C} \) is joined to \( \mathrm{M} \) and produced to a point \( \mathrm{D} \) such that \( \mathrm{DM}=\mathrm{CM} \). Point \( \mathrm{D} \) is joined to point \( \mathrm{B} \) (see Fig. 7.23). Show that:(i) \( \triangle \mathrm{AMC} \equiv \triangle \mathrm{BMD} \)(ii) \( \angle \mathrm{DBC} \) is a right angle.(iii) \( \triangle \mathrm{DBC} \equiv \triangle \mathrm{ACB} \)(iv) \( \mathrm{CM}=\frac{1}{2} \mathrm{AB} \)"
- If \( \mathrm{B} \) is the mid point of \( \overline{\mathrm{AC}} \) and \( \mathrm{C} \) is the mid point of \( \overline{\mathrm{BD}} \), where \( \mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D} \) lie on a straight line, say why \( \mathrm{AB}=\mathrm{CD} \)?
- Choose the correct answer from the given four options:If in two triangles \( \mathrm{DEF} \) and \( \mathrm{PQR}, \angle \mathrm{D}=\angle \mathrm{Q} \) and \( \angle \mathrm{R}=\angle \mathrm{E} \), then which of the following is not true?(A) \( \frac{\mathrm{EF}}{\mathrm{PR}}=\frac{\mathrm{DF}}{\mathrm{PQ}} \)(B) \( \frac{\mathrm{DE}}{\mathrm{PQ}}=\frac{\mathrm{EF}}{\mathrm{RP}} \)(C) \( \frac{\mathrm{DE}}{\mathrm{QR}}=\frac{\mathrm{DF}}{\mathrm{PQ}} \)(D) \( \frac{E F}{R P}=\frac{D E}{Q R} \)
- \( \mathrm{D} \) and \( \mathrm{E} \) are points on sides \( \mathrm{AB} \) and \( \mathrm{AC} \) respectively of \( \triangle \mathrm{ABC} \) such that ar \( (\mathrm{DBC})=\operatorname{ar}(\mathrm{EBC}) \). Prove that $DE\|BC$.
- In figure below, \( \mathrm{BD} \) and \( \mathrm{CE} \) intersect each other at the point \( \mathrm{P} \). Is \( \triangle \mathrm{PBC} \sim \triangle \mathrm{PDE} \) ? Why?"
- In a quadrilateral \( \mathrm{ABCD}, \angle \mathrm{A}+\angle \mathrm{D}=90^{\circ} \). Prove that \( \mathrm{AC}^{2}+\mathrm{BD}^{2}=\mathrm{AD}^{2}+\mathrm{BC}^{2} \) [Hint: Produce \( \mathrm{AB} \) and DC to meet at E]
- \( \mathrm{O} \) is the point of intersection of the diagonals \( \mathrm{AC} \) and \( \mathrm{BD} \) of a trapezium \( \mathrm{ABCD} \) with \( \mathrm{AB} \| \mathrm{DC} \). Through \( \mathrm{O} \), a line segment \( \mathrm{PQ} \) is drawn parallel to \( \mathrm{AB} \) meeting \( \mathrm{AD} \) in \( \mathrm{P} \) and \( \mathrm{BC} \) in \( \mathrm{Q} \). Prove that \( \mathrm{PO}=\mathrm{QO} \).
- It is given that \( \triangle \mathrm{ABC} \sim \Delta \mathrm{EDF} \) such that \( \mathrm{AB}=5 \mathrm{~cm} \), \( \mathrm{AC}=7 \mathrm{~cm}, \mathrm{DF}=15 \mathrm{~cm} \) and \( \mathrm{DE}=12 \mathrm{~cm} \). Find the lengths of the remaining sides of the triangles.
Kickstart Your Career
Get certified by completing the course
Get Started