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Pythagoras Theorem

aIn Fig., D i...

Question

(a)
In Fig., $D$ is a point on hypotenuse $A C$ of $Δ A B C,$ such that $B D ⊥ A C, D M ⊥ B C and D N ⊥ A B .$
Prove that : $(i) D M^{2} = D N . M C$

(b)
In Fig., $D$ is a point on hypotentenuse $A C$ of $Δ A B C,$ such that
$B D ⊥ A C, D M ⊥ B C and D N ⊥ A B . Prove that :$

(ii) $D N^{2} = D M \times A N$

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Solution

(a)
Analyzing figure $D M B N$
In $Δ A B C,$
$D N ⊥ N B and B M ⊥ N B$
So, $D N | | B M$
$D M ⊥ B M and B N ⊥ B M$
So, $D M | | B N$
In figure $D M B N$
$D N | | B M$
$D M | | B N$
and all interior angles are $90^{\circ}$
Hence, $D M B N$ is a rectangle.
Therefore, $B M = D M$

In $Δ B M D, Δ D M C & Δ A B C$
In $Δ B M D,$
$∠ D M B + ∠ B D M + ∠ D B M = 180^{\circ}$
$∠ B D M + ∠ D B M = 90^{\circ} . . . (1)$
In $Δ D M C,$
$∠ C D M + ∠ M C D + ∠ D M C = 180^{\circ}$
$∠ C D M + ∠ M C D = 90^{\circ} . . . (2)$
In $Δ A B C$
We know, $B D ⊥ A C ∴ ∠ B D M + ∠ M D C = 90^{\circ} . . . (3)$

Substituting ralues in equations.
From (1) and (3), we get
$∠ B D M + ∠ D B M = ∠ B D M + ∠ M D C$
$\Rightarrow ∠ D B M = ∠ M D C . . . . . . . . . (4)$
similarly, $∠ B D M ∠ M C D . . . . . . . (5)$

Checking for $A A A$ in $Δ B M D$ $Δ D M C$
In $Δ B M D$ and $Δ D M C,$
$∠ B M D = ∠ D M C = 90^{\circ}$
$∠ D B M = ∠ M D C . . .$ From (4)
$∠ B D M = ∠ M C D$ .... From (5)
By $A A A$ dimilarity criteria
$Δ B M D \sim Δ D M C$

Using property of similar triangles $Δ B M D$ and $Δ D M C$
$\Rightarrow \frac{D M}{B M} = \frac{M C}{D M}$
$∵ B M = N D$
$\Rightarrow \frac{D M}{D N} = \frac{M C}{D M}$
$D M^{2} = D N . M C$
Hence proved.

(b)
Analyzing figure $D M B N$
In $Δ A B C$
$D N ⊥ N B and B M ⊥ N B$
So, $D N | | B M$
$D M ⊥ B M and B N ⊥ B M$
So, $D M | | B N$
In figure $D M B N$
$D N | | B M$
$D M | | B N$
and all inter interior angles are $90^{\circ}$
hence, $D M B N$ is a rectangle.

Finding out relation between the angles
$a n g l e N D B + ∠ M D B = 90^{\circ} . . . . . . . (1)$
$∠ A D N + ∠ D A N = 90^{\circ} . . . . . . . (2)$
$∠ A D N + ∠ N D B = 90^{\circ} . . . . . . . (3)$

Therefore, from (1) and (3)
$∠ M D B = ∠ A D N$
Therefore, from (2) and (3)
$∠ D A N = ∠ N D B$

Using $A A$ similarity in $Δ D N B$ and $Δ A N D$
$∠ D N B = ∠ A N D$
$∠ N D B = ∠ D A N$
Hence,
$Δ D N B \sim Δ A N D$
Using $C P S T$
$\Rightarrow \frac{A N}{D N} = \frac{D N}{N B}$

$D N^{2} = D M \times A N$
Proved.

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Similar questions

D

A C

△ A B C

B D ⊥ A C, D M ⊥ B C

D N ⊥ A B .

D M^{2} = D N . M C

D N^{2} = D M . A N

D

A C

△ A B C

B D ⊥ A C

D M ⊥ B C

D N ⊥ A B

{D M}^{2} = D N . M C

{D N}^{2} = D M . A N

In the given figure, D is a point on hypotenuse AC of ΔABC, DM ⊥ BC and DN ⊥ AB, Prove that:

(i) DM² = DN.MC

(ii) DN² = DM.AN

$I n Δ ABC, D is the midpoint of BC. if D L ⊥ A B a n d D M ⊥ A C such that DL = DM, prove that AB = AC.$

D M = D N

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