In the given figure, AC = AE. Show that: i) CP = EP ii) BP = DP - Noon Academy

In the given figure, AC = AE. Show that: i) CP = EP ii) BP = DP

Class 10 | Exercise 18C | ICSE | Maths | Selina | Tangents and Intersecting Chords

In the given figure, AC = AE. Show that: i) CP = EP ii) BP = DP

Selina Solutions Concise Class 10 Maths Chapter 18 ex. 18(C) - 20

Solution:

In

$\vartriangle ADC\text{ }and\text{ }\vartriangle ABE$

$\angle ACD\text{ }=\angle AEB~$

[Angles in the same segment]

$AC\text{ }=\text{ }AE$

[Given]

$\angle A\text{ }=\angle A$

[Common]

Hence,

$\vartriangle ADC\cong \vartriangle ABE\text{ }by\text{ }ASA$

postulate

So, by C.P.C.T we have

$AB\text{ }=\text{ }AD$

But,

$AC\text{ }=\text{ }AE$

[Given]

So,

$AC\text{ }-\text{ }AB\text{ }=\text{ }AE\text{ }-\text{ }AD$

$BC\text{ }=\text{ }DE$

In

$\vartriangle BPC\text{ }and\text{ }\vartriangle DPE$

$\angle C\text{ }=\angle E~$

[Angles in the same segment]

So,

$BC\text{ }=\text{ }DE$

$\angle CBP\text{ }=\angle CDE$

[Angles in the same segment]

Hence,

$\vartriangle BPC\cong \vartriangle DPE\text{ }by\text{ }ASA$

postulate

By C.P.C.T we have

$BP\text{ }=\text{ }DP\text{ }and\text{ }CP\text{ }=\text{ }PE$

i

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