In the fig., AB = CB, AB = CD and EF bisects BD at G. Prove that G is mid-point of EF.
Answers
Step-by-step explanation:
Given;
AB = CB
AB = CD
EF bisects BD at G. So, DG = BG;
To Prove : G is the midpoint of EF or GE = GF.
Construction: Join A and C with a straight line.
Proof : In △ BAC;
∠CAB = ∠ACB (angles opposite to equal sides are equal) (i)
In △ GDC and △GBC;
GD = GB. (given)
DC = BC. (given)
GC = GC. (common)
Therefore, △ GDC ≅ △GBC. (by SSS criteria of congruence)
Hence, ∠DGC = ∠BGC (C.P.C.T)
∠DCG = ∠BCG (C.P.C.T) (ii)
In △ GBC and △GBA;
∠CAB = ∠ACB (from eq. (i))
∠AGB = ∠DGC (vertically opposute angle)
AB = BC (given)
Therefore, △ GBC ≅ △GBA (by AAS criteria of congruence)
Hence, GC = AG (C.P.C.T) (iii)
∠BAG = ∠BCG (C.P.C.T) (iv)
From equation (ii) and (iv);
∠DCG = ∠BAG. (v)
∠BAG = ∠GAF (vi)
∠DCG = ∠GCE. (vii)
Now, in △CGE and △AGF;
∠GAF = ∠GCE. (from equation (v), (vi) and (vii))
AG = GC. (from equation (iii))
∠AGF = ∠CGE. (vertically opposite angle)
Therefore, △CGE ≅ △AGF (by ASA criteria of congruence)
Hence, GE = GF. (by C.P.C.T)
That's all.