Question

only legends can answer this question​

Answer 1

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

In ∆ABC, D is any point on BC such that ∠BAC = ∠ADC.

$\large\underline{\sf{To\:prove - }}$

$\boxed{ \rm{ {CA}^{2} = CB \times CD}}$

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

Given that,

In ∆ABC, D is any point on BC such that ∠BAC = ∠ADC.

$\rm :\longmapsto\:In \: \triangle \: ACD \: and \: \triangle \: BCA$

$\red{\rm :\longmapsto\:\angle BAC \: = \: \angle \: ADC \: \: \: \{given \}}$

$\red{\rm :\longmapsto\:\angle BCA \: = \: \angle \: ACD \: \: \: \{common \}}$

$\bf\implies \triangle \: ACD \sim \triangle \: BCA \: \: \{By \: AA \: Similarity \}$

Therefore,

$\rm :\longmapsto\:\dfrac{CA}{CB} = \dfrac{CD}{CA}$

$\bf\implies \:\boxed{ \bf{ {CA}^{2} = CB \times CD}}$

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.

Answer 2

Answer:

\large\underline{\sf{Given- }}

Given−

In ∆ABC, D is any point on BC such that ∠BAC = ∠ADC.

\large\underline{\sf{To\:prove - }}

Toprove−

\boxed{ \rm{ {CA}^{2} = CB \times CD}}

CA

2

=CB×CD

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

Solution−

Given that,

In ∆ABC, D is any point on BC such that ∠BAC = ∠ADC.

\rm :\longmapsto\:In \: \triangle \: ACD \: and \: \triangle \: BCA:⟼In△ACDand△BCA

\red{\rm :\longmapsto\:\angle BAC \: = \: \angle \: ADC \: \: \: \{given \}}:⟼∠BAC=∠ADC{given}

\red{\rm :\longmapsto\:\angle BCA \: = \: \angle \: ACD \: \: \: \{common \}}:⟼∠BCA=∠ACD{common}

\bf\implies \triangle \: ACD \sim \triangle \: BCA \: \: \{By \: AA \: Similarity \}⟹△ACD∼△BCA{ByAASimilarity}

Therefore,

\rm :\longmapsto\:\dfrac{CA}{CB} = \dfrac{CD}{CA} :⟼

CB

CA

=

CA

CD

\bf\implies \:\boxed{ \bf{ {CA}^{2} = CB \times CD}}⟹

CA

2

=CB×CD

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.