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SSS Criteria for Congruency

In right tria...

Question

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 (see the given figure). Show that:
$Δ D B C ≅ Δ A C B$

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Solution

$I n Δ A M C a n d Δ B M D A M = B M (M i s m i d p o i n t o f A B) ∠ A M C = ∠ B M D (v e r t i c a l l y o p p o s i t e a n g l e s) C M = D M (g i v e n) ∴ Δ A M C ≅ Δ B M D (b y S A S c o n g r u e n c e r u l e) ∴ A C = B D (b y C P C T) A n d ∠ A C M = ∠ B D M (b y C P C T) W e h a v e ∠ A C M = ∠ B D M B u t A C M a n d B D M a r e a l t e r n a t e i n t e r i o r a n g l e s S i n c e a l t e r n a t e a n g l e s a r e e q u a l . H e n c e, w e c a n s a y t h a t D B | | A C \Rightarrow ∠ D B C + ∠ A C B = 180^{\circ} (c o - i n t e r i o r a n g l e s) \Rightarrow ∠ D B C + 90^{\circ} = 180^{\circ} (∠ A C B = 90^{\circ}) \Rightarrow ∠ D B C = 180^{\circ} - 90^{\circ} \Rightarrow ∠ D B C = 90^{\circ} \Rightarrow D B = A C (B y C P C T) \dots (i) I n Δ D B C a n d Δ A C B D B = A C (F r o m (i)) B C = B C (C o m m o n) ∠ D B C = ∠ A C B = 90^{\circ} ∴ Δ D B C ≅ Δ A C B b y S A S c o n g r u e n c e$

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Q. 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 (see the given figure). Show that:

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Q. Question 8 (iii)
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 (see the given figure). Show that:

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