Mathematics
In the given figure, AC = AE.
Show that :
(i) CP = EP
(ii) BP = DP
Circles
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Answer
(i) In ∆ADC and ∆ABE,
⇒ ∠ACD = ∠AEB [Angles in the same segment are equal]
⇒ AC = AE [Given]
⇒ ∠A = ∠A [Common]
Hence, ∆ADC ≅ ∆ABE by ASA axiom.
So, by C.P.C.T we have
⇒ AD = AB …………..(1)
Given,
⇒ AE = AC ………….(2)
Subtracting equation (1) from (2), we get :
⇒ AE - AD = AC - AB
⇒ DE = BC
In ∆BPC and ∆DPE,
⇒ ∠C = ∠E [Angles in the same segment are equal]
⇒ BC = DE [Proved above]
⇒ ∠CBP = ∠PDE [Angles in the same segment are equal]
Hence, ∆BPC ≅ ∆DPE by ASA axiom.
So, by C.P.C.T we have
⇒ CP = EP
Hence, proved that CP = EP.
(ii) Proved above,
∆BPC ≅ ∆DPE
∴ BP = DP [By C.P.C.T]
Hence, proved that BP = DP.
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