CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

Two circles touch each other internally at a point $$P$$. A chord $$AB$$ of the bigger circle intersects the other circle in $$C$$ and $$D$$.Prove that : $$\angle{CPA}=\angle {DPB}$$
1205450_67a105f7278c4559a002015c3fdb11c4.png


Solution

Draw a tangent TS at P to the circles given
Since TPS is the tangent, PD is the chord.
$$\therefore\angle{PAB}=\angle{BPS}....(1)$$   [Angles in alternate segment]
Similarly,
$$\angle{PCD}=\angle{DPS}....(2)$$
Subtracting (1) and (2)
$$\angle{PCD}-\angle{PAB}=\angle{DPS}-\angle{BPS}$$
But in $$\triangle{PAC}$$
Exterior $$\angle{PCD}=\angle{PAB}+\angle{CPA}$$
$$\therefore \angle{PAB}+\angle{CPA}-\angle{PAB}=\angle{DPS}-\angle{BPS}$$
$$\Rightarrow \angle{CPA}=\angle{DPB}$$

1110287_1205450_ans_6852e0229b6449ef93fe1e7716fcd6b9.png

Mathematics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image