Question
In a parallelogram ABCD, the bisector of $\angle\text{A}$ also bisects BC at X. Find AB : AD.
✓
Answer
Parallelogram ABCD is given as follows: 
We have AX bisects $\angle\text{A}$ bisecting BC at X. That is, BX = CX We need to find AB : AD Since, AX is the bisector $\angle\text{A}$ That is,$\angle1=\frac{1}{2}\angle\text{A}\ ....(\text{i})$
Also, ABCD is a parallelogram Therefore, AD || BC and AB intersects them$\angle\text{A}+\angle\text{B}=180^\circ$
$\angle\text{B}=180^\circ-\angle\text{A}\ ....(\text{ii})$
In $\triangle\text{ABX},$ by angle sum property of a triangle:$\angle1+\angle2+\angle\text{B}=180^\circ$
From (i) and (ii), we get:$\frac{1}{2}\angle\text{A}+\angle2+180^\circ-\angle\text{A}=180^\circ$
$\angle2-\frac{1}{2}\angle\text{A}=0$
$\angle2=\frac{1}{2}\angle\text{A}\ ....(\text{iii})$
From (i) and (iii),we get:$\angle1=\angle2$
Sides opposite to equal angles are equal. Therefore, BX = AB 2BX = 2AB As X is the mid point of BC. Therefore, BC = 2AB Also, ABCD is a parallelogram, then, BC = AD AD = 2AB Thus, AB : AD = AB : 2AB AB : AD = 1 : 2 Hence the ratio of AB : AD is 1 : 2.
We have AX bisects $\angle\text{A}$ bisecting BC at X. That is, BX = CX We need to find AB : AD Since, AX is the bisector $\angle\text{A}$ That is,$\angle1=\frac{1}{2}\angle\text{A}\ ....(\text{i})$Also, ABCD is a parallelogram Therefore, AD || BC and AB intersects them$\angle\text{A}+\angle\text{B}=180^\circ$
$\angle\text{B}=180^\circ-\angle\text{A}\ ....(\text{ii})$
In $\triangle\text{ABX},$ by angle sum property of a triangle:$\angle1+\angle2+\angle\text{B}=180^\circ$
From (i) and (ii), we get:$\frac{1}{2}\angle\text{A}+\angle2+180^\circ-\angle\text{A}=180^\circ$
$\angle2-\frac{1}{2}\angle\text{A}=0$
$\angle2=\frac{1}{2}\angle\text{A}\ ....(\text{iii})$
From (i) and (iii),we get:$\angle1=\angle2$
Sides opposite to equal angles are equal. Therefore, BX = AB 2BX = 2AB As X is the mid point of BC. Therefore, BC = 2AB Also, ABCD is a parallelogram, then, BC = AD AD = 2AB Thus, AB : AD = AB : 2AB AB : AD = 1 : 2 Hence the ratio of AB : AD is 1 : 2.
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 missing frequency (p) for the following distribution whose mean is 7.68
|
x:
|
3
|
5
|
7
|
9
|
11
|
13
|
|
f:
|
6
|
8
|
15
|
p
|
8
|
4
|
In the given figure, AB || CD || EF. Find the value of x.
The water bills (in rupees) of 32 houses in a certain street for the period 1.198 to 31.398 are given below:
56, 43, 32, 38, 56, 24, 68, 85, 52, 47, 35, 58, 63, 74, 27, 84, 69, 35, 44, 75, 55, 30, 54, 65, 45, 67, 95, 72, 43, 65, 35, 69.
Tabulate the data and present the data as a cumulative frequency table using 70 - 79 as one of the class intervals.
56, 43, 32, 38, 56, 24, 68, 85, 52, 47, 35, 58, 63, 74, 27, 84, 69, 35, 44, 75, 55, 30, 54, 65, 45, 67, 95, 72, 43, 65, 35, 69.
Tabulate the data and present the data as a cumulative frequency table using 70 - 79 as one of the class intervals.
Factorize:$\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)-4\Big(\text{x}+\frac{1}{\text{x}}\Big)+6$
→In the adjoining figure, ABCD is a parallelogram in which AB is produced to E so that BE = AB. Prove that ED bisects BC.
In two right triangles one side an acute angle of one are equal to the corresponding side and angle of the other. Prove that the triangles are congruent.
In the given figure, line RP || line MS and line DK is their transversal. ∠DHP = 85°. Find the measures of following angles.
i. ∠RHD
ii. ∠PHG
iii. ∠HGS
iv. ∠MGK
A cylindrical vessel, without lid, has to be tin-coated on its both sides. If the radius of the base is 70 cm and its height is 1.4 m , calculate the cost of tin-coating at the rate of $₹ 3.50$ per $1000 cm^2$.
→If $\text{x}+\frac{1}{\text{x}}=5,$ find the value of $\text{x}^3+\frac{1}{\text{x}^3}.$
→The ratio of two numbers is 5 : 7. If 40 is added in each number, then the ratio becomes
25 : 31, Find the numbers.
25 : 31, Find the numbers.