D is a point on the side BC of ABC such that - ADC- BAC- Prove that CA-CD-CB-CA or CA2-CB- CD

In ABC and DAC , we have ∠ ADC=∠ BAC and ∠ C=∠ C ∴ By AA axiom of similarity ABC∼ DAC ⇒ AB/DA=BC/AC=AC/DC ⇒ ...

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Standard X

Mathematics

AAA Similarity

D is a point ...

Question

$D$ is a point on the side $B C$ of $△ A B C$ such that $∠ A D C = ∠ B A C .$ Prove that $\frac{C A}{C D} = \frac{C B}{C A}$ or $C A^{2} = C B \times C D$

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Solution

In $△ A B C$ and $△ D A C$ , we have
$∠ A D C = ∠ B A C$ and $∠ C = ∠ C$
$∴$ By AA-axiom of similarity
$△ A B C \sim △ D A C$
$\Rightarrow$ $\frac{A B}{D A} = \frac{B C}{A C} = \frac{A C}{D C}$
$\Rightarrow$ $\frac{C B}{C A} = \frac{C A}{C D}$ ....Hence, proved

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D is a point on the side BC of △ABC such that ∠ADC=∠BAC. Prove that CACD=CBCA or CA2=CB×CD

In △ABC and △DAC, we have∠ADC=∠BAC and ∠C=∠C∴ By AA-axiom of similarity△ABC∼△DAC⇒ ABDA=BCAC=ACDC⇒ CBCA=CACD ....Hence, proved

$D$ is a point on the side $B C$ of $△ A B C$ such that $∠ A D C = ∠ B A C .$ Prove that $\frac{C A}{C D} = \frac{C B}{C A}$ or $C A^{2} = C B \times C D$

In $△ A B C$ and $△ D A C$ , we have
$∠ A D C = ∠ B A C$ and $∠ C = ∠ C$
$∴$ By AA-axiom of similarity
$△ A B C \sim △ D A C$
$\Rightarrow$ $\frac{A B}{D A} = \frac{B C}{A C} = \frac{A C}{D C}$
$\Rightarrow$ $\frac{C B}{C A} = \frac{C A}{C D}$ ....Hence, proved