Question
In given figure, it is given that $\text{AB = CF , EF = BD}$ and $\angle AFE =\angle CBD$. Prove that $\triangle AFE \cong \triangle CBD$.
✓
Answer
In triangles $\text{AFE}$ and $\text{CBD} ($ in above shown figure$)$, we have
$\text{AB=CF}($Given $)$
Adding $\text{BF}$ on both the sides, we get$:-$
$\text{AB + BF = CF + BF}$
or, $\text{AF = BC}$
Now in triangles $\text{AFE}$ and $\text{CBD}$, we have $\text{AF= CB} ($Proved above$)$
$\angle AFE =\angle CBD($ Given $)$
and $\text{EF = BD}($ Given$)$
So, according to $\text{SAS}$ congruency criteria of triangles;
$\triangle AFE \cong \triangle CBD$
Hence, proved.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Find the following products:
$(3 x-4 y+5 z)\left(9 x^2+16 y^2+25 z^2+12 x y-15 z x+20 y z\right)$
→$(3 x-4 y+5 z)\left(9 x^2+16 y^2+25 z^2+12 x y-15 z x+20 y z\right)$
On a certain day, the temperature in a city was recorded as under.
Illustrate the data by a bar graph.
→|
Time
|
$5 a.m.$ | $8 a.m.$ |
$11 a.m.$
|
$3 p.m.$
|
$6 p.m.$ |
|
Temperature (in °C)
|
$20$
|
$24$
|
$26$
|
$22$
|
$18$ |
Show that $\mathrm{p}-1$ is a factor of $\mathrm{p}^{10}-1$ and also of $\mathrm{p}^{11}-1$.
→ABC is an isosceles triangle with $AB = AC$ and $D$ is a point on $BC$ such that $\text{AD}\perp\text{BC}$ (see figure). To prove that $\angle\text{BAD} = \angle\text{CAD},$ a student proceeded as follows:
In $\triangle\text{ABD}$ and $\triangle\text{ACD},$
$\text{AB}=\text{AC}$
$\angle\text{B}=\angle\text{C}$
$\angle\text{ADM}=\angle\text{ADC}$
$\therefore\triangle\text{ABD}\cong\triangle\text{ADC}$
$\angle\text{BAD}=\angle\text{CAD}$
What is the defect in the above arguments?
→In $\triangle\text{ABD}$ and $\triangle\text{ACD},$
$\text{AB}=\text{AC}$
$\angle\text{B}=\angle\text{C}$
$\angle\text{ADM}=\angle\text{ADC}$
$\therefore\triangle\text{ABD}\cong\triangle\text{ADC}$
$\angle\text{BAD}=\angle\text{CAD}$
What is the defect in the above arguments?
In the given figure, $BAD\ ||\ EF$, $\angle\text{AEF}=55^\circ$ and $\angle\text{ACB}=25^\circ,$ find $\angle\text{ABC}.$ 
→The table given below shows the marks obtained by $30$ students in a test.
Out of these students, one is chosen at random. What is the probability that the marks of the chosen student:
$i.$ Are $30$ or less?
$ii.$ Are $31$ or more?
$iii.$ Lie in the interval $21 - 30$?
→| Marks $($Class interval$)$ |
$1 - 10$ | $11 - 20$ | $21 - 30$ | $31 - 40$ | $41 - 50$ |
| Number of students $($Frequency$)$ |
$7$ | $10$ | $6$ | $4$ | $3$ |
$i.$ Are $30$ or less?
$ii.$ Are $31$ or more?
$iii.$ Lie in the interval $21 - 30$?
Factorise: $\text{x}^2-2\text{x}+\frac{7}{16}$
→An angle is equal to five times its complement. Determine its measure.
A Joker's cap is in the form of a right circular cone of base radius $7 \ cm$ and height $24 \ cm$. Find the area of the sheet required to make $10$ such caps.
→$\text{ABC}$ is a triangle. $D$ is a point on $\text{AB}$ such that $AD =\frac{1}{4} AB$ and $E$ is a point of $AC$ such that $AE =\frac{1}{4} AC$. Prove that $DE =\frac{1}{4} BC$.
→