In ∆ A B C , A D ⊥ B C and A D 2 = B D × C D . Prove that A B 2 + A C 2 = B D + C D 2 .

In △ A B D A B 2 = A D 2 + B D 2 △ A D C A C 2 = A D 2 + C D 2 (i) + (ii) A B 2 + A C 2 = 2 A D 2 + B D 2 + C D 2 Put A D 2 = B D × C D A B 2 + A C 2 = 2 B D . C D + B D 2 + C D 2 A B 2 + A C 2 = B D 2 + C D 2 + 2 B D · C D A B 2 + A C 2 = ( B D + C D ) 2 Hence, proved.

In ∆ A B C , ∠ A B C = 90 ° , B D ⊥ A C . If B D = 8 c m , A D = 4 c m , find C D and A B .

Given that, In ∆ A B C , ∠ A B C = 90 ° , B D ⊥ A C . If B D = 8 c m , A D = 4 c m , Then, in ∆ A B D By Pythagoras theorem, A B 2 = A D 2 + B D 2 A B 2 = 4 2 + 8 2 A B 2 = 16 + 64 A B = 80 A B = 4 5 c m Now in ∆ B D C let ∠ B C D = x then, ∠ D B C = 90 ° - x (By angle sum property) Then in ∆ A B D , ∠ A B D = x and ∠ B A D = 90 ° - x (By angle sum property) Now in ∆ A D B and ∆ B D C ∠ A B D = ∠ B C D = x ∠ B A D = ∠ D B C = 90 ° - x ∠ A D B = ∠ B D C (Both 90 ° ) ∴ ∆ A D B ~ ∆ B D C (By A A A ) Then, C D B D = B D A D (Two triangles are said to be similar if their corresponding angles are congruent and the corresponding sides are in proportion) C D 8 = 8 4 C D = 16 c m Therefore, C D = 16 c m and A B = 4 5 c m .

Draw a chord of length 6 c m in a circle of radius 5 c m . Measure and write the distance of the chord from the centre of the circle.

As the line drawn perpendicular from the centre of the chord bisects the chord, Therefore, A C = 3 c m , and O A = 5 c m which is the radius. A O C is a right-angled triangle and O C is the distance between the chord and the centre. As O A 2 = A C 2 + O C 2 , O C = O A 2 - A C 2 ∴ O C = 5 2 - 3 2 O C = 25 - 9 = 16 O C = 4 c m The distance of the chord from the centre of the circle is 4 c m .

Prove that ''if in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar''.

This theorem can be proved by taking two triangles A B C and D E F such that ∠ A = ∠ D , ∠ B = ∠ E , ∠ C = ∠ F . Cut D P = A B , D Q = A C and join P Q . In △ A B C and △ D P Q , A B = D P (by construction) A C = D Q (by construction) ∠ A = ∠ D (given) ⇒ △ A B C ≅ △ D P Q (By SAS criterion) ⇒ ∠ B = ∠ P (By CPCT) But ∠ B = ∠ E (given) ⇒ ∠ E = ∠ P But ∠ P and ∠ E are corresponding angles, therefore, P Q ∥ E F . Therefore, D P P E = D Q Q F i.e. P E D P = Q F D Q Adding 1 to both sides gives, P E D P + 1 = Q F D Q + 1 ⇒ P E + D P D P = Q F + D Q D Q ⇒ D E D P = D F D Q ⇒ D E A B = D F A C ⇒ A B D E = A C D F Similarly, A B D E = B C E F Therefore, A B D E = A C D F = B C E F Since all 3 sides are in proportion. ∴ △ A B C ~ △ D E F . Hence, proved.

In the △ A B D , C is a point on B D such that B C : C D = 1 : 2 , and △ A B C is an equilateral triangle. Then prove that A D 2 = 7 A C 2 .

Given: A triangle A B D , in which C is a point on side B D such that B C : C D = 1 : 2 and △ A B C is an equilateral triangle. We need to prove, A D 2 = 7 A C 2 . Draw A P ⊥ B D , so A P bisect the side B C . ⇒ B P = P C = 1 2 B C = 1 2 A C In a ∆ A B C , A B = B C = A C ⇒ B D = 3 B C = 3 A C According to the Pythagoras theorem, In a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides. Therefore, in a triangle A C P , ∠ P = 90 ° A C 2 = A P 2 + C P 2 ⇒ A P 2 = A C 2 - C P 2 ⇒ A P 2 = A C 2 - 1 2 A C 2 = 3 4 A C 2 . . . . . . . . . i Now, applying Pythagoras theorem to the ∆ A D P , ⇒ A D 2 = A P 2 + P D 2 ⇒ A D 2 = 3 4 A C 2 + B D - B P 2 ⇒ A D 2 = 3 4 A C 2 + 3 A C - 1 2 A C 2 ⇒ A D 2 = 3 4 A C 2 + 25 4 A C 2 ⇒ A D 2 = 28 4 A C 2 ⇒ A D 2 = 7 A C 2 Hence, proved.