CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

$$D$$ is a point on the side $$BC$$ of a triangle $$ABC$$ such that $$\angle ADC = \angle BAC$$. Show that $$CA^2 = CB.CD$$.


Solution

In $$\triangle ADC$$ and $$\triangle BAC$$

$$\angle ADC=\angle BAC$$ (Given)

$$\angle C$$ is Common

$$\therefore$$ by AA Criterion of Similarity, $$\triangle ADC$$ $$\sim$$ $$\triangle BAC$$

$$\Rightarrow \dfrac{AD}{BA}=\dfrac{DC}{AC}=\dfrac{AC}{BC}$$

$$\Rightarrow \dfrac{DC}{AC}=\dfrac{AC}{BC}$$

$$\therefore$$ $$CA^2=CB.CD$$

494378_465441_ans.png

Mathematics
RS Agarwal
Standard X

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
Same exercise questions
View More


similar_icon
People also searched for
View More



footer-image