In Fig. 5.10, if $ \mathrm{AC}=\mathrm{BD} $, then prove that $ \mathrm{AB}=\mathrm{CD} $.
"
Given:
$AC=BD$.
To do:
We have to prove that $AB=CD$.
Solution:
Given,
$AC=BD$
According to the Figure,
We get,
$AC=AB+BC$
$BD=BC+CD$
This implies,
$AB+BC=BC+CD$....(i) (Since, $AC=BD$
According to Euclid's Axiom,
When equals are subtracted from equals, reminders are also equal.
Therefore,
Let us subtract $BC$ from both sides of (i)
We get,
$AB+BC-BC=BC+CD-BC$
This implies,
$AB=CD$
Hence proved.
- Related Articles
- In Fig. 6.29, if \( \mathrm{AB}\|\mathrm{CD}, \mathrm{CD}\| \mathrm{EF} \) and \( y: z=3: 7 \), find \( x \)."\n
- Find \( \mathrm{x} \) if \( \mathrm{AB}\|\mathrm{CD}\| \mathrm{EF} \)."\n
- In Fig. 7.21, \( \mathrm{AC}=\mathrm{AE}, \mathrm{AB}=\mathrm{AD} \) and \( \angle \mathrm{BAD}=\angle \mathrm{EAC} \). Show that \( \mathrm{BC}=\mathrm{DE} \)."\n
- In figure below, \( l \| \mathrm{m} \) and line segments \( \mathrm{AB}, \mathrm{CD} \) and \( \mathrm{EF} \) are concurrent at point \( \mathrm{P} \).Prove that \( \frac{\mathrm{AE}}{\mathrm{BF}}=\frac{\mathrm{AC}}{\mathrm{BD}}=\frac{\mathrm{CE}}{\mathrm{FD}} \)."
- In Fig. 6.30, if \( \mathrm{AB} \| \mathrm{CD} \), EF \( \perp \mathrm{CD} \) and \( \angle \mathrm{GED}=126^{\circ} \), find \( \angle \mathrm{AGE}, \angle \mathrm{GEF} \) and \( \angle \mathrm{FGE} \)."\n
- In Fig. 6.15, \( \angle \mathrm{PQR}=\angle \mathrm{PRQ} \), then prove that \( \angle \mathrm{PQS}=\angle \mathrm{PRT} \)"\n
- \( \mathrm{AD} \) and \( \mathrm{BC} \) are equal perpendiculars to a line segment \( \mathrm{AB} \) (see Fig. 7.18). Show that \( \mathrm{CD} \) bisects \( \mathrm{AB} \)."\n
- In Fig. 6.28, find the values of \( x \) and \( y \) and then show that \( \mathrm{AB}=\mathrm{CD} \)."\n
- Diagonals \( \mathrm{AC} \) and \( \mathrm{BD} \) of a trapezium \( \mathrm{ABCD} \) with \( \mathrm{AB} \| \mathrm{DC} \) intersect each other at \( \mathrm{O} \). Prove that ar \( (\mathrm{AOD})=\operatorname{ar}(\mathrm{BOC}) \).
- In figure below, if \( \mathrm{PQRS} \) is a parallelogram and \( \mathrm{AB} \| \mathrm{PS} \), then prove that \( \mathrm{OC} \| \mathrm{SR} \)."
- In figure below, diagonals \( \mathrm{AC} \) and \( \mathrm{BD} \) of quadrilateral \( \mathrm{ABCD} \) intersect at \( \mathrm{O} \) such that \( \mathrm{OB}=\mathrm{OD} \).If \( \mathrm{AB}=\mathrm{CD} \), then show that:(i) \( \operatorname{ar}(\mathrm{DOC})=\operatorname{ar}(\mathrm{AOB}) \)(ii) \( \operatorname{ar}(\mathrm{DCB})=\operatorname{ar}(\mathrm{ACB}) \)(iii) DA\| \( \mathrm{CB} \) or \( \mathrm{ABCD} \) is a parallelogram.[Hint : From D and B, draw perpendiculars to AC.]"\n
- In quadrilateral \( A C B D \),\( \mathrm{AC}=\mathrm{AD} \) and \( \mathrm{AB} \) bisects \( \angle \mathrm{A} \) (see Fig. 7.16). Show that \( \triangle \mathrm{ABC} \cong \triangle \mathrm{ABD} \).What can you say about \( \mathrm{BC} \) and \( \mathrm{BD} \) ?"64391"\n
- In Fig. 7.48, sides \( \mathrm{AB} \) and \( \mathrm{AC} \) of \( \triangle \mathrm{ABC} \) are extended to points \( \mathrm{P} \) and \( \mathrm{Q} \) respectively. Also, \( \angle \mathrm{PBC}\mathrm{AB} \)."\n
- In Fig. 6.32, if \( \mathrm{AB} \| \mathrm{CD}, \angle \mathrm{APQ}=50^{\circ} \) and \( \angle \mathrm{PRD}=127^{\circ} \), find \( x \) and \( y \)."\n
- \( \triangle \mathrm{ABC} \) is an isosceles triangle in which \( \mathrm{AB}=\mathrm{AC} \). Side \( \mathrm{BA} \) is produced to \( \mathrm{D} \) such that \( \mathrm{AD}=\mathrm{AB} \) (see Fig. 7.34). Show that \( \angle \mathrm{BCD} \) is a right angle."\n
Kickstart Your Career
Get certified by completing the course
Get Started
Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies