Question

In Fig. 14, prove that triangle ABD = triangle CDB.​

Answer 1

Answer:

Step-by-step explanation:

in triangle ABD and  triangle CDB.​

BD is common

∠ABD=∠CBD ( ATl angles)

∠DAB=∠DCB ( VOA of equal angles)

So ΔCDB ≅ ΔABD

Answer 2

Answer:

GIVEN:- <HAG=<FCE (GIVEN)

<DAB=<BCE (<HAG=<FCE)

IN ∆ABD & ∆CDB

AB//CD

<ABD=<CDB (ALTERNATIVE INTERIOR ANGELES)

<DAB=<BCE (<HAG=<FCE)

BD=BD ( COMMON SIDE)

∆ABD=∆CDB (BY AAS)