Question

ABC is an isosceles triangle whose ∠C is right angle. If D is any point on AB, then let us prove that, AD² + DB² = 2CD². ​

Answer 1

Step-by-step explanation:

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Answer 2

$\large\underline{\sf{Given \: appropriate \:Question - }}$

ABC is an isosceles triangle whose ∠C is right angle. If D is any point on AB such that D is midpoint of AB, then let us prove that, AD² + DB² = 2CD².

$\large\underline{\sf{Solution-}}$

Given that,

➢ ABC is an isosceles triangle in which ∠C is right angle.

⇛ AC = BC

➢ D is mid point of AB

⇛ BD = AD

➢ Consider, Δ ABC

By Pythagoras theorem,

$\boxed{ \rm \: {AB}^{2} = {AC}^{2} + {BC}^{2}} - - - (1)$

➢ Now, In Δ ADC and Δ CDB

AC = BC               [Given]

CD = CD              [Common]

AD = DB              [D is the mid point]

⇒ ΔADC ≅ ΔBDC             [ By SSS Congruency Rule ]

⇒ ∠ADC = ∠BDC           [ CPCT ]

Also, ∠ADC + ∠BDC = 180°      [Linear Pair]

⇒ ∠ADC + ∠ADC = 180°

⇒ 2 ∠ADC = 180°

⇒ ∠ADC = 90°

⇒ ∠BDC = 90°

Now, Consider Δ BDC

By Pythagoras theorem

$\boxed{ \rm \: {BC}^{2} = {BD}^{2} + {DC}^{2}} - - - - (2)$

Now, Consider Δ ADC

By Pythagoras theorem,

$\boxed{ \rm \: {AC}^{2} = {AD}^{2} + {DC}^{2}} - - - - (3)$

Now, On adding equation (2) and (3), we get

$\rm :\longmapsto\: {BC}^{2} + {AC}^{2} = {AD}^{2} + {BD}^{2} + 2 {DC}^{2}$

$\rm :\longmapsto\: {AB}^{2} = {AD}^{2} + {BD}^{2} + 2 {DC}^{2}$

$\red{\bigg \{ \sf \because \:using \: equation \: (1) \bigg \}}$

$\rm :\longmapsto\: {(2AD)}^{2} = {AD}^{2} + {BD}^{2} + 2 {DC}^{2}$

$\red{\bigg \{ \because \:AB = 2AD = 2BD \bigg \}}$

$\rm :\longmapsto\: {4AD}^{2} = {AD}^{2} + {BD}^{2} + 2 {DC}^{2}$

$\rm :\longmapsto\: {2AD}^{2} + 2 {AD}^{2} = {AD}^{2} + {BD}^{2} + 2 {DC}^{2}$

$\rm :\longmapsto\: {AD}^{2} + 2 {BD}^{2} = {BD}^{2} + 2 {DC}^{2}$

$\red{\bigg \{ \because \:AD = BD \bigg \}}$

$\rm :\longmapsto\: {AD}^{2} + {BD}^{2} = 2 {DC}^{2}$

$\large{{\boxed{\bf{Hence, Proved}}}}$

Additional Information

1. Pythagoras Theorem :-

This theorem states that : In a right-angled triangle, the square of the longest side is equal to sum of the squares of remaining sides.

2. Converse of Pythagoras Theorem :-

This theorem states that : If the square of the longest side is equal to sum of the squares of remaining two sides, angle opposite to longest side is right angle.

3. Area Ratio Theorem :-

This theorem states that :- The ratio of the area of two similar triangles is equal to the ratio of the squares of corresponding sides.

4. Basic Proportionality Theorem

If a line is drawn parallel to one side of a triangle, intersects the other two lines in distinct points, then the other two sides are divided in the same ratio.