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
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
In right triangle $ \mathrm{ABC} $, right angled at $ \mathrm{C}, \mathrm{M} $ is the mid-point of hypotenuse $ \mathrm{AB} $. $ \mathrm{C} $ is joined to $ \mathrm{M} $ and produced to a point $ \mathrm{D} $ such that $ \mathrm{DM}=\mathrm{CM} $. Point $ \mathrm{D} $ is joined to point $ \mathrm{B} $ (see Fig. 7.23). Show that:
(i) $ \triangle \mathrm{AMC} \equiv \triangle \mathrm{BMD} $
(ii) $ \angle \mathrm{DBC} $ is a right angle.
(iii) $ \triangle \mathrm{DBC} \equiv \triangle \mathrm{ACB} $
(iv) $ \mathrm{CM}=\frac{1}{2} \mathrm{AB} $
"
Given:
In right triangle $ABC$, right angled at $C$, $M$ is the mid-point of hypotenuse $AB$. $C$ is joined to $M$ and produced to a point $D$ such that $DM=CM$. Point $D$ is joined to point $B$.
To do:
We have to show the given questions.
Solution:
(i) Let us consider $\triangle AMC$ and $\triangle BMD$,
We know that,
According to Rule of Side-Angle-Side Congruence:
Triangles are said to be congruent if any pair of corresponding sides and their included angles are equal in both triangles.
Given,
$CM=DM$ and
$M$ is the midpoint of the hypotenuse $AB$
This implies,
$AM=BM$
We also know that,
The vertically opposite angles are always equal,
This implies,
$\angle CMA=\angle DMB$
Therefore,
By SAS congruence,
$\triangle AMC \cong \triangle BMD$.
(ii) We also know,
From corresponding parts of congruent triangles: If two triangles are congruent, all of their corresponding angles and sides must be equal.
This implies,
$\angle ACM=\angle BDM$
We know that,
if the alternate interior angles were equal, then the two lines were parallel.
This implies,
$AC \parallel BD$
We also know that, the sum of the interior angles is equal to $180^o$
This implies,
$\angle ACB+\angle DBC=180^o$
$90^o+\angle DBC=180^o$ (given, $\triangle ABC$ is right angled at $c$)
This implies,
$\angle DBC=90^o$
(iii) Let us consider $\triangle DBC$ and $\triangle ACB$
We know that,
According to Rule of Side-Angle-Side Congruence:
Triangles are said to be congruent if any pair of corresponding sides and their included angles are equal in both triangles.
$BC=CB$ (common side)
Since,$\angle ACB$ and $DBC$ are right angles to each other we get,
$\angle ACB=\angle DBC$
We also know,
From corresponding parts of congruent triangles: If two triangles are congruent, all of their corresponding sides must be equal.
Therefore,
$DB=AC$.
Therefore,
$\triangle DBC \cong ACB$.
(iv) Since $\triangle DBC \cong ACB$
We get,
$DM=AB$
We also have $M$ as the mid point
This implies,
$DM=CM=AM=BM$
Therefore,
$DM+CM=BM+AM$
$CM+CM=AB$
This implies,
$CM=\frac{1}{2}AB$.
28 Views
