Question

In fig. $ ABC $ is a right triangle right angled at $ A $ . $ BCED $ , $ BCED $ , $ ACFG $ and $ ABMN $ are square on the sides $ BC $ , $ CA $ and $ AB $ respectively. Line segment $ AX \bot DE $ meets $ BC $ at $ Y $ . Show that:

 $ \Delta MBC \cong \Delta ABD $
 $ {\rm{ar}}\left( {BYXD} \right) = 2{\rm{ar}}\left( {MBC} \right) $
 $ {\rm{ar}}\left( {BYXD} \right) = {\rm{ar}}\left( {ABMN} \right) $
 $ \Delta FCB \cong \Delta ACE $
 $ {\rm{ar}}\left( {CYXE} \right) = 2{\rm{ar}}\left( {FCB} \right) $
 $ {\rm{ar}}\left( {CYXE} \right) = {\rm{ar}}\left( {ACFG} \right) $
 $ {\rm{ar}}\left( {BCED} \right) = {\rm{ar}}\left( {ABMN} \right) + {\rm{ar}}\left( {ACFG} \right) $

Answer 1

Hint: To show that two triangles are congruent use some models like SAS, SSS, AAA. To show that areas are equal, try to make the left hand side equal to the area of some part and the right hand side equal to the area of the same part. Then there will be some relation from which you can equate both.

Complete step-by-step answer:
In the figure, $ ABC $ is a right angle triangle right angled a $ A $ .

 $ BCED $ , $ ACFG $ and $ ABMN $ are squares on the sides $ BC $ , $ CA $ and $ AB $ respectively.
Line segment $ AX \bot DE $ meets $ BC $ at $ Y $ .

 (i)
From the given expression, we know that ABMN is a square then we have,
 $ MB = AB $ ……(1)
We also know that sides of the square are equal so,
  $ BC = BD $ ……(2)
Angles of the square are equal so $ \angle MBA = \angle CBD $ since each angle is $ 90^\circ $ in square.
On adding $ \angle ABC $ for the above equation on both sides we get,
 $ \begin{array}{c}
\angle MBA + \angle ABC = \angle CBD + \angle ABC\\
\angle MBC = \angle ABD
\end{array} $ …. (3)
We know the SAS congruence rule. On equating the equations (1), (2) and (3), we get the value as,
 $ \Delta MBC \cong \Delta ABD $
Therefore, it is proved that $ \Delta MBC \cong \Delta ABD $ .

 (ii)
On observing the diagram, we can say that,
 $ {\rm{ar}}\left( {ABD} \right){\rm{ = }}\dfrac{1}{2} \times {\rm{ar}}\left( {BYDX} \right) $ ……(4)
We have proven this in the above part that,
 $ {\rm{ar}}\left( {ABD} \right) = {\rm{ar}}\left( {MBC} \right) $ ……..(5)
On comparing equations (4) and (5) we get,
 $ {\rm{ar}}\left( {MBC} \right) = \dfrac{1}{2} \times {\rm{ar}}\left( {BYDX} \right) $
Hence, it proves that $ {\rm{ar}}\left( {BYDX} \right) = 2{\rm{ar}}\left( {MBC} \right) $ .

 (iii)
Since, we know that from above relations,
 $ {\rm{ar}}\left( {BYXD} \right) = 2{\rm{ar}}\left( {ABD} \right) $ .......(6)
On applying the property of squares, we have,
 $ {\rm{ar}}\left( {ABMN} \right) = 2{\rm{ar}}\left( {MBC} \right) = 2{\rm{ar}}\left( {ABD} \right) $ ........(7)
On equating the equations, (6) and (7), we have,
 $ {\rm{ar}}\left( {BXYD} \right) = {\rm{ar}}\left( {ABMN} \right) $
Hence, it proves that $ {\rm{ar}}\left( {BXYD} \right) = {\rm{ar}}\left( {ABMN} \right) $ .

 (iv)
In $ \Delta FCB $ and $ \Delta ACE $ ,
 $ FC = AC $ (sides of the squares are equal) ........(8)
 $ CB = CE $ (sides of the squares are equal) ........(9)
 $ \angle FCA = \angle BCE $ (each angles in square are equal and the angle is $ 90^\circ $ )
Adding $ \angle ACB $ both sides for the above equation we get,
 $ \begin{array}{c}
\angle FCA + \angle ACB = \angle BCE + \angle ACB\\
\angle FCB = \angle ACE
\end{array} $ ......(10)
So, by comparing equations (8), (9) and (10) we can say that the triangles FCB and ACE are congruent by SAS congruence.
 $ \Delta FCB \cong \Delta ACE $
Hence, it proves that $ \Delta FCB \cong \Delta ACE $ .

 (v)
By the diagram we can conclude that area of CYXE is equal to twice the area of triangle ACE, that is,
 $ {\rm{ar}}\left( {CYXE} \right) = 2{\rm{ar}}\left( {ACE} \right) $ ......(11)

We also know by SAS congruence that triangles $ \Delta FCB \cong \Delta ACE $ have equal area.
 $ {\rm{ar}}\left( {ACE} \right) = {\rm{ar}}\left( {FCB} \right) $ .......(12)
So on comparing equations (11) and (12) we get,
 $ {\rm{ar}}\left( {CYXE} \right) = 2{\rm{ar}}\left( {FCB} \right) $
Hence, it proves that $ {\rm{ar}}\left( {CYXE} \right) = 2{\rm{ar}}\left( {FCB} \right) $ .

(vi)
We know by previous proofs that,
 $ {\rm{ar}}\left( {CYXE} \right) = 2{\rm{ar}}\left( {FCB} \right) $ ........(13)
 $ {\rm{ar}}\left( {ACFG} \right) = 2{\rm{ar}}\left( {FCB} \right) $ .......(14)
By equating the above equations (13) and (14) we get,
 $ {\rm{ar}}\left( {CYXE} \right) = {\rm{ar}}\left( {ACFG} \right) $

(vii)
From proof of part (iii) and (vi), we get the relations,
 $ {\rm{ar}}\left( {BYXD} \right) = {\rm{ar}}\left( {ABMN} \right) $
 $ {\rm{ar}}\left( {CYXE} \right) = {\rm{ar}}\left( {ACFG} \right) $
On adding both the equations, we get,
 $ {\rm{ar}}\left( {BYXD} \right) + {\rm{ar}}\left( {CYXE} \right) = {\rm{ar}}\left( {ABMN} \right) + {\rm{ar}}\left( {ACFG} \right) $
Since, we know that,
 $ {\rm{ar}}\left( {BYXD} \right) + {\rm{ar}}\left( {CYXE} \right) = {\rm{ar}}\left( {BCED} \right) $
On substituting the value in the above relation, we will get,
 $ {\rm{ar}}\left( {BCED} \right) = {\rm{ar}}\left( {ABMN} \right) + {\rm{ar}}\left( {ACFG} \right) $
Hence, it is proved that $ {\rm{ar}}\left( {BCED} \right) = {\rm{ar}}\left( {ABMN} \right) + {\rm{ar}}\left( {ACFG} \right) $ .

Note: In this problem, the proof of the fifth part can also be proved by using Pythagoras theorem where $ A $ is right angled. Using the Pythagoras theorem, the fifth result is obvious. The theorem says that the square of the hypotenuse side is equal to the sum of square of base and square of height.