Question
$ABCD$ is a quadrilateral in which $AB || DC$ and $AD = BC$. Prove that $\angle\text{A}=\angle\text{B}$ and $\angle\text{C}=\angle\text{D}.$
✓
Answer
$ABCD$ is a quadrilaateral such that $AB || DC$ and $AD = BC$
Construction Extend $AB$ to $E$ and draw a line $CE$ parallet to $AD.$ proos since, $AD || CE$ and transverrsal $AE$ cuts then at $A$ and $E$ repectively.
$\therefore\angle\text{A}+\angle\text{E}=180^\circ [$since, sum of cointerios angles is $180^\circ ]$
$\Rightarrow\ \angle\text{A}=180^\circ-\angle\text{E}\ ...(\text{i})$
So, quadrilateral $AECD$ is a parallelogram.
$\Rightarrow\ \text{AD}=\text{CE}\Rightarrow\text{BC}=\text{CE}$ [$\because\ \text{AD}=\text{BC}$ given]
Now, in $\Delta\text{BCE}$
$\text{CE}=\text{BC}$ [proved above]
$\Rightarrow\ \angle\text{CBE}=\angle\text{CEB}$ [opposite angles of equal side are equal]
$\Rightarrow\ 180^\circ-\angle\text{B}=\angle\text{E}$
$[\because\angle\text{B}+\angle\text{CBE}=180^\circ]$
$\Rightarrow\ 180^\circ-\angle\text{E}=\angle\text{B}\ ...(\text{ii})$
From Eqs. $(i)$ and $(ii) \angle\text{A}=\angle\text{B}$ H
ence proved.
Construction Extend $AB$ to $E$ and draw a line $CE$ parallet to $AD.$ proos since, $AD || CE$ and transverrsal $AE$ cuts then at $A$ and $E$ repectively.
$\therefore\angle\text{A}+\angle\text{E}=180^\circ [$since, sum of cointerios angles is $180^\circ ]$
$\Rightarrow\ \angle\text{A}=180^\circ-\angle\text{E}\ ...(\text{i})$
So, quadrilateral $AECD$ is a parallelogram.
$\Rightarrow\ \text{AD}=\text{CE}\Rightarrow\text{BC}=\text{CE}$ [$\because\ \text{AD}=\text{BC}$ given]
Now, in $\Delta\text{BCE}$
$\text{CE}=\text{BC}$ [proved above]
$\Rightarrow\ \angle\text{CBE}=\angle\text{CEB}$ [opposite angles of equal side are equal]
$\Rightarrow\ 180^\circ-\angle\text{B}=\angle\text{E}$
$[\because\angle\text{B}+\angle\text{CBE}=180^\circ]$
$\Rightarrow\ 180^\circ-\angle\text{E}=\angle\text{B}\ ...(\text{ii})$
From Eqs. $(i)$ and $(ii) \angle\text{A}=\angle\text{B}$ H
ence 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
Draw the graph of the following linear equations in two variables: $x - y = 2$
→In a $\triangle ABC$, if $L$ and $M$ are points on $AB$ and $AC$ respectively such that $LM \| BC$. Prove that:
$i. \text{ar}(\triangle\text{LCM})=\text{ar}(\triangle\text{LBM})$
$ii. \text{ar}(\triangle\text{LBC})=\text{ar}(\triangle\text{MBC})$
$iii. \text{ar}(\triangle\text{ABM})=\text{ar}(\triangle\text{ACL})$
$iv. \text{ar}(\triangle\text{LOB})=\text{ar}(\triangle\text{MOC})$
→$i. \text{ar}(\triangle\text{LCM})=\text{ar}(\triangle\text{LBM})$
$ii. \text{ar}(\triangle\text{LBC})=\text{ar}(\triangle\text{MBC})$
$iii. \text{ar}(\triangle\text{ABM})=\text{ar}(\triangle\text{ACL})$
$iv. \text{ar}(\triangle\text{LOB})=\text{ar}(\triangle\text{MOC})$
A rectangular tank is $80\ m$ long and $25\ m$ broad. Water flows into it through a pipe whose cross-section is $25\ cm^2$, at the rate of $16\ km$ per hour. How much the level of the water rises in the tank in $45$ minutes?
→In figure, $AB = AC$ and $CP \| BA$ and $AP$ is the bisector of exterior $\angle\text{CAD}$ of $\triangle\text{ABC}.$ Prove that:
$i. \angle\text{PAC}=\angle\text{BCA}.$
$ii. \text{ABCP}$ is a parallelogram.
→$i. \angle\text{PAC}=\angle\text{BCA}.$
$ii. \text{ABCP}$ is a parallelogram.
Two chords AB and CD of lengths 5cm and 11cm respectively of a circle are parallel to each other and are opposite side of its centre. If the distance between AB and CD is 6cm, find the radius of the circle.
$PQR$ is a triangle in which $PQ = PR$ and is any point on the side $PQ$. Through $S$, a line is drawn parallel to $QR$ and intersecting $PR$ at $T$. Prove that $PS = PT.$
→A circular park of radius 40m is situated in a colony. Three boys Ankur, Amit and Anand are sitting at equal distance on its boundary each having a toy telephone in his hands to talk to each other. Find the length of the string of each phone.
Find the area of a triangle whose sides are respectively $9\ cm, 12\ cm$ and $15\ cm.$
→Find the volume, the lateral surface area and the total surface area of the cuboid whose dimensions are: Length $= 26\ m,$ breadth $= 14\ m$ and height $= 6.5\ m$
→It costs ₹ 2200 to paint the inner curved surface of a cylindrical vessel 10m deep. If the cost of painting is at the rate of ₹ 20 per m2, find:
→- Inner curved surface area of the vessel.
- Radius of the base.
- Capacity of the vessel.