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Intersection between Secants

In the given ...

Question

In the given figure, AC = AE.

Show that :

(i) CP = EP

(ii) BP = DP.

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Solution

In $△ A D C a n d △ A B E, ∠ A C D = ∠ A E B (a n g l e s i n t h e s a m e s e g m e n t)$

$A C = A E$

∠A=∠A

$∴ △ A D C ≅ △ A B E (A S A)$

$\to A B = A D b u t A C = A E$

$\to A C - A B = A E - A D$

$\to B C = D E$

In $△ B P C a n d △ D P E, ∠ C = ∠ E (a n g l e s i n t h e s a m e s e g m e n t)$

$B C = D E$

∠CBP=∠CDE

$∴ △ B P C ≅ △ D P E (A S A)$

BP = DP and CP = PE (cpct)

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Similar questions

In the given figure, AC = AE. Select the statements that are true.

A C

B D

P

\frac{A P}{C P} = \frac{B P}{D P}

△ A B P \sim △ C D P

A B C D

B P Q

B P = B C

D Q

C P

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∠ C D Q

In the given figure, AP = BP and CP = DP. Which of the following statements are true?

In the figure, given alongside, CP bisects angle ACB.

Show that DP bisects angle ADB.

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