Question

in the given figure triangle ABC.DPllAC and DEllAP where P lies on BC.prove that BE/EC=BC/CP

Answer 1

In ΔABC, we have,

∠B=2∠C or ∠B=2y where ∠C=y

AD is the bisector of ∠BAC. So, Let ∠BAD=∠CAD=x

Let BP be the bisector of ∠ABC. Join PD

In ΔBPC, we have

∠CBP = ∠BCP = y ⇒ BP = PC ... (1)

Now, in ΔABP and ΔDCP, we have

∠ABP = ∠DCP = y

AB = DC [Given]

and, BP = PC [Using (1)]

So, by SAS congruence criterion, we have

Δ ABP ≅ Δ DCP

Therefore

∠BAP = ∠ CDP = 2x and AP = DP ,

So in Δ APD, AP=DP

=> ∠ADP = ∠DAP = x

In ΔABD, we have

∠ADC = ∠ABD + BAD ⇒ 3x= 2y + x ⇒ x = y

In ΔABC, we have

∠A + ∠B + ∠C = 180°

⇒ 2x + 2y + y = 180°

⇒ 5x = 180°

⇒ x = 36°

Hence, ∠BAC = 2x = 72°