Question

In the given figure, D and E are points on AB and AC respectively. AE and CD intersect at P such that AP = CP. If ∠BAE = ∠BCD, prove that DBDE is isosceles.

Answer 1

Join DE and AC
In ΔAPD = ΔEPC,
∠DAP = ∠ECP       ....(∵ ∠BAE = ∠BCD)
AP = CP                 ....(given)
∠APD = ∠EPC      ....(vertically opposite angles)
∴ ΔAPD ≅ ΔEPC    ....(By ASA Congruence critetion)
⇒ AD = EC            ....(c.p.c.t)
In ΔAPC,
AP = CP                 ....(given)
⇒ ∠PAC = ∠PCA  ....(angles opposite to two equal sides are equal)
Now, ∠BAC = ∠BCD and ∠PAC = ∠PCA
⇒ ∠BAC =∠BCA
⇒  BC = BA            ....(sides opposites to two equal sides are equal)
⇒ BE + EC = BD + DA
⇒ BE = BD            ....(∵ EC = DA)
⇒ ∠BDE = ∠BED ....(angles opposite to two equal sides are equal)
⇒ ΔBDE is an isosceles triangle.