8. In the given figure, AD is bisector of ∠BAC and ∠CPD = ∠BPD. Prove that ΔCAP ≌ ΔBAP.
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Given : AD is the bisector of ∠BAC
∠CPD = ∠BPD
To Find : Prove that ΔCAP ≅ ΔBAP
Solution:
AD is the bisector of ∠BAC
=> ∠CAP = ∠BAP
∠APC + ∠CPD = 180° linear pair
∠APB + ∠BPD = 180° linear pair
=> ∠APC + ∠CPD = ∠APB + ∠BPD
∠CPD = ∠BPD
=> ∠APC = ∠APB
in ΔCAP and ΔBAP
∠CAP = ∠BAP
AP = AP common
∠APC = ∠APB
=> ΔCAP ≅ ΔBAP ( ASA)
QED hence proved
mid-point theorem :
line joining the mid-point of two sides of a triangle is equal to half the length of the third side and parallel
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