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
A conical tent is 10m high and the radius of it base is 24m. Find the slant height of the tent. If the cost of 1m2 canvas is Rs. 70, find the cost of canvas required for the tent.
Multiply:
(9x2 + 25y2 + 15xy + 12x - 20y + 16) by (3x - 5y + 4)
(9x2 + 25y2 + 15xy + 12x - 20y + 16) by (3x - 5y + 4)
In the adjoining figure, ABCD is a quadrilateral in which diagonal BD = 14 cm. If $\text{AL}\perp\text{BD}$ and $\text{CM}\perp\text{BD}$such that AL = 8cm and CM = 6cm, find the area of quadrilateral ABCD.
→In fig lines XY and MN intersect at O If $\angle $POY = 90° and a:b = 2:3 find $\angle $c.
→Find the volume of a sphere whose surface area is 154cm2. $\big(\text{Take}\ \pi=\frac{22}{7}\big).$
→Prove that:
$\Big(\frac{64}{125}\Big)^{-\frac{2}{3}}+\frac{1}{\Big(\frac{256}{625}\Big)^\frac{1}{4}}+\Big(\frac{\sqrt{25}}{\sqrt[3]{64}}\Big)^0=\frac{61}{16}$
→$\Big(\frac{64}{125}\Big)^{-\frac{2}{3}}+\frac{1}{\Big(\frac{256}{625}\Big)^\frac{1}{4}}+\Big(\frac{\sqrt{25}}{\sqrt[3]{64}}\Big)^0=\frac{61}{16}$
Use suitable identity to find the product:
$\left( {x + 4} \right)\left( {x + 10} \right)$
→$\left( {x + 4} \right)\left( {x + 10} \right)$
ABCD is a cyclic qudrilateral in which:
$ \angle\text{DBC}=80^\circ$ and $\angle\text{BAC}=40^\circ.$ Find $\angle\text{BCD}.$
→$ \angle\text{DBC}=80^\circ$ and $\angle\text{BAC}=40^\circ.$ Find $\angle\text{BCD}.$
If the diagonals of a parallelogram are equal, then show that it is a rectangle.
It being given that $\sqrt{3}=1.732,\sqrt{5}=2.236,\sqrt{6}=2.449$ and $\sqrt{10}=3.162,$ find to three places of decimal, the value of the following:
$\frac{1}{4\sqrt{3}-3\sqrt{5}}$
→$\frac{1}{4\sqrt{3}-3\sqrt{5}}$