Question
In the given figure, $AD \| BE \| CF.$
Prove that area $(\triangle AEC) =$ area $(\triangle DBF)$
Prove that area $(\triangle AEC) =$ area $(\triangle DBF)$
✓
Answer
We know that the area of triangles on the same base and between the same parallel lines are equal. Consider $ABED$ quadrilateral; $AD \| BE.$
With the common base, $BE$ and between $AD$ and $BE$ parallel lines,
we have
Area of $\triangle ABE =$ Area of $\triangle BDE$ Similarly, in $\text{BEFC}$ quadrilateral,
$BE \| CF$ With common base $BC$ and between $BE$ and $CF$ parallel lines,
we have
Area of $\triangle BEC = $Area of $\triangle BEF$ Adding both equations, we have
Area of $\triangle ABE +$ Area of $\triangle BEC =$ Area of $\triangle BEF +$ Area of $\triangle BDE$
$\Rightarrow $ Area of $AEC =$ Area of $DBF$
Hence Proved.
With the common base, $BE$ and between $AD$ and $BE$ parallel lines,
we have
Area of $\triangle ABE =$ Area of $\triangle BDE$ Similarly, in $\text{BEFC}$ quadrilateral,
$BE \| CF$ With common base $BC$ and between $BE$ and $CF$ parallel lines,
we have
Area of $\triangle BEC = $Area of $\triangle BEF$ Adding both equations, we have
Area of $\triangle ABE +$ Area of $\triangle BEC =$ Area of $\triangle BEF +$ Area of $\triangle BDE$
$\Rightarrow $ Area of $AEC =$ Area of $DBF$
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
Evaluate : $\frac{\cos 3 A+2 \cos 4 A}{\sin 3 A+2 \sin 4 A}$, when $A=15^{\circ}$.
→Factorise : $9a^2+ 3a - 8b - 64b^2$
→If $\log \left(\frac{a+b}{2}\right)=\frac{1}{2}(\log a+\log b)$, show that $\frac{1}{2}(a+b)=\sqrt{a b}$.
→A circle of radius 2.5 cm has a chord of length 4.8 cm . Find the distance of the chord from the centre of the circle.
In the adjoining figure, $AC > AB$ and AD is the bisector of $\angle A$. Show that : $\angle ADC >\angle ADB$.
→Find the value of $'a\ '$ and $'b\ '$ if $a. (a + 2,5 + b) = (1, 6),b. (2a + b, a - 2b) = (7, 6)$
→The area of a square garden is equal to the area of a rectangular plot of length $160\ m$ and width $40\ m.$ Calculate the cost of fencing the square garden at $Rs.12$ per m.
→$10%$ of $x + 20%$ of $y = 24,3x - y = 20$
→Find the area of a ring whose outer and inner radii are 19 cm and 16 cm respectively.
A rectangular plot 30 m long and 18 m wide is to be covered with grass leaving 2.5 m all around it. Find the area to be laid with grass.