Question
ABCD is a quadrilaeter in which AB = BC and AD = CD. Show that BD bisects both the angles ABC and ADC.
✓
Answer
In $\triangle\text{ABC}$ and $\triangle\text{CBD},$ we have
$\text{AB}=\text{BC}$
$\text{AD}=\text{CD}$
$\text{BD}=\text{BD}$
$\therefore\text{ABC}\cong\triangle\text{CBD}$
$\Rightarrow\angle\text{1}=\angle\text{2}$
$\angle3=\angle4$
Hence, BD bises both the angle ABC and ADC.
$\text{AB}=\text{BC}$$\text{AD}=\text{CD}$
$\text{BD}=\text{BD}$
$\therefore\text{ABC}\cong\triangle\text{CBD}$
$\Rightarrow\angle\text{1}=\angle\text{2}$
$\angle3=\angle4$
Hence, BD bises both the angle ABC and ADC.
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
In the given figure, PQRS is an isosceles trapezium. Find x and y.
Factorise:
$\text{x}^2+\frac{1}{\text{x}^2}-3$
→$\text{x}^2+\frac{1}{\text{x}^2}-3$
What are:
- Primary data,
- Secondary data?
A rectangular plot is given for constructing a house, having a measurement of $40m$ long and $15m$ in the front. According to the laws, a minimum of $3m$ wide space should be left in the front and back each and $2m$ wide space on each of the other sides. Find the largest area where house can be constructed.
→Express the following decimals in the form $\frac{\text{p}}{\text{q}}:$
$0.\bar{4}$
→$0.\bar{4}$
Solve the following equations for x:
$2^{\text{2x}}-2^{\text{x}+3}+2^4=0$
→$2^{\text{2x}}-2^{\text{x}+3}+2^4=0$
Show that:$\Big(\frac{\text{a}^{\text{x}+1}}{\text{a}^{\text{y}+1}}\Big)^{\text{x}+\text{y}}\Big(\frac{\text{a}^{\text{y}+2}}{\text{a}^{\text{z}+2}}\Big)^{\text{y}+\text{z}}\Big(\frac{\text{a}^{\text{z}+3}}{\text{a}^{\text{x}+3}}\Big)^{\text{z}+\text{x}}=1$
→A conical tent is $10\ m$ high and the radius of its base is $24\ m$. Find
→- Slant height of the tent.
- Cost of the canvas required to make the tent, if the cost of $1 m^2$ canvas is $Rs. 70.$
Write in decimal form and say what kind of decimal expansion: $\frac{{36}}{{100}}$
→Factorise:
$\sqrt{5}\text{x}^2+2\text{x}-3\sqrt{5}$
→$\sqrt{5}\text{x}^2+2\text{x}-3\sqrt{5}$